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«•>«< 





































THE LATEST WO^K 

ON 



481 PAGES AND 13 FOLDING PLATES. $3.50. 


The Stresses in Bridge and Roof Trusses, Arched 
Ribs apd Suspension Bridges, ar\d Cantilevers. 

Prepared for the Department of Civil Engineering at the Rensselaer Polytechnic Institute, 

By WILLIAM H. BURR, C.E., 

PROFESSOR OF CIVIL ENGINEERING IN COLUMBIA COLLEGE IN THE CITY OF NEW YORK; MEMBER OF THE 

AMERICAN SOCIETY OF CIVIL ENGINEERS. 


EIGHTH REVISED AND ENLARGED EDITION. 


NEW YORK: 

JOHN WILE^Y & SONS. 

1893. 

COMPLETE , CONCISE AND PRACTICAL. DESIGNED FOR TECHNICAL 

STUDENTS AND ENGINEERS. 


* * * “ There are 124 pages on Ordinary Bridge and Roof Trusses, 99 on Draw- 
Bridges, 50 on Arched Ribs, and 34 on Suspension Bridges. At the end is a chapter on the 
details of construction. * * * Both algebraic and graphic methods are presented, while 

the prominence is given to the former. A single algebraic method is not rigidly followed, 
but often two or more are given, so that one may serve as a check on the other. Methods 
are everywhere made subordinate to principles, and for a student no better plan can be fol¬ 
lowed. * * * The section on Draw-Bridges will prove of especial value to advanced 

students and bridge engineers. It is, indeed, the only presentation of that important subject 
which is at all suitable to put before a class. There are doubtless many instructors of engi¬ 
neering who will welcome such a carefully prepared and exhaustive book on the determination 
of stresses.— Railroad Gazette. 


This work has been carefully prepared with two objects in view; 
one, to meet the wants of the practicing engineer and bridge builder, 
and the other, to provide students who intend pursuing an advanced 
course of study in civil engineering, such a course of instruction in the 
determination of stresses in structures of iron, steel and wood, as will 
fit them to discharge, in a practical and efficient manner, professional 
duties connected with such work. 

During the past few years, the science of iron and steel construction 
has made rapid progress, and the loose, “ rule of thumb ” methods, 
which were formerly permissible, no longer find place among advanced 






























“The Latest Engineering Experience and Tests.” 


THE 

ELASTICITY AND RESISTANCE 


ON THE 



By WM. H. BURR, C.E., 

PROFESSOR OF CIVIL ENGINEERING IN COLUMBIA COLLEGE IN THE CITY OF NEW YORK J MEMBER OF THE 

AMERICAN SOCIETY OF CIVIL ENGINEERS. 


A PRACTICAL TREATISE ON THE RESISTANCE OR STRENGTH OF 

MATERIALS. 


THIRD REVISED AND ENLARGED EDITION. 


8vo, 755 Pages, Cloth, $5.00. 


PUBLISHED AND FOR SALE BY 

JOHN WILEY & SONS, 53 East 1 Oth St., New York. 

Will be mailed and prepaid on the receipt of the price. 


This work has been the outgrowth of lectures on the elasticity and 
resistance of materials, given by the author to succeeding classes of students 
in the department of Civil Engineering at the Rensselaer Polytechnic Insti¬ 
tute. Although those lectures form the basis of the work, they have 
been considerably elaborated and extended so as to include many ad¬ 
ditional details of importance to the practical engineer. 

It is believed that the author’s practical experience as an engineer, 
and familiarity with the methods and needs of advanced technical instruc¬ 
tion, has enabled him to satisfactorily comprehend in this book those 
things which are essential both to the technical student and engineer. 

The rational or theoretical part covers less than one-third the 
volume. In this part the student or investigator will find a concise treat¬ 
ment of the philosophical basis, so to speak, of the resistance of materials. 
Instead of making the theories of torsion, flexure, etc., pure assumptions, 
as is usually done, those theories are shown to be simply logical expres¬ 
sions of Hooke’s law, combined with the elementary principles of statics, 
applied to the particular manifestations of external forces or loads. The 
subject is thus put upon a common sense foundation and not started on 
a mere empiricism or conventionality. 



































A COURSE 


THE STRESSES 

IN 


BRIDGE AND ROOF TRUSSES, ARCHED RIBS 
AND SUSPENSION BRIDGES, 


PREPARED FOR THE DEPARTMENT OF CIVIL ENGINEERING AT THE 
RENSSELAER POLYTECHNIC INSTITUTE. 


ijf' 


By WILLIAM H. BURR, C.E., 

•» 

PROFESSOR OF CIVIL ENGINEERING IN COLUMBIA COLLEGE IN THE CITY OF NEW YORK ; 
MEMBER OF THE AMERICAN SOCIETY OF CIVIL ENGINEERS. 


EIGHTH EDITION REVISED AND ENLARGED. 


THIRD THOUSAND. 


5 > 
) ) ) 


NEW YORK: 

JOHN WILEY & SONS. 

1893. 







Copyright, 

1886, 

By Wm. H. Burr. 


T ransfer 

Engineers School Li by. 

June 29,1931 



NSW YORK: J. J. LITTLE St CO.. PRINTERS, 
10 TO 20 ASTOR PLACE. 




PREFACE TO THE EIGHTH EDITION. 


The eighth edition of this work has been considerably 
enlarged by a very material extension of Arts. 35, 36, 37, and 
3 ya, on Swing Bridges, which have been entirely rewritten. 
In this revision, I am under much indebtedness to Mr. 
Henry W. Hodge, C.E., for the computations involved in 
the numerical work of the two actual cases of swing bridge 
spans in Arts. 36 and 37. The general theory of this class 
of structures, as first completely given in the first edition of 
this work, has proved to be so well adapted to the rather 
complicated requirements of the case that it is now almost 
universally adopted in this country. The essential improve¬ 
ments and simplifications presented in this new matter make 
the labor of computation but little more than that required 
for non-continuous spans. 


New York City, 
July, 1893. 


W. H. B. 




PREFACE TO THIRD EDITION. 


SINCE the publication of the first edition of this book 
engineering practice in iron and steel construction, especially 
in the department of bridge building, has made very material 
progress. The distribution of metal in pin structures has 
been considerably modified so as to produce concentrations 
in larger members; but chiefly the treatment of moving 
loads has experienced such a radical transformation as to 
bring it to a thoroughly rational basis. Hence, portions of 
the book as originally written have been cancelled and re¬ 
placed by entirely new matter, so amplified and extended as 
to bring the work in all its details abreast of the best practice 
of the present day. 

My indebtedness to the published papers of Prof. H. T. 
Eddy, of the University of Cincinnati, on the arched rib, will 
be evident to any reader even slightly acquainted with his 
valuable work entitled “ Researches in Graphical Statics.” 

Certain matters are of such common occurrence in the fol¬ 
lowing pages that it may conduce to clearness to mention 
them h£re. 

The word “ ton ” signifies a ton of 2,000 pounds, unless it 
is otherwise specifically stated. 

The word “ stress ” means the force acting in any member 
of a structure, while “strain” is the distortion which accom¬ 
panies the stress. 

The sign -j- indicates a tensile stress, and the sign —, a 
compressive one. 

Unless otherwise stated, the stress in any member of a 
structure will be represented by inclosing with a parenthesis 
the letter or letters which belong to it in the diagrams or 
plates. Thus (A B), or (a), signifies “stress in the member 
A B,” or “stress in the member a .” 



VI 


PREFACE. 


As a matter of convenience to those who may be familiar 
with the first and second editions, it is well to state that 
pages 19 to 60, Arts. 72, 74 and 85 are entirely new. Portions 
of pages at several other places in the book have also been 
re-written, but it is unnecessary to name them here. 

For convenience in swing bridge computations, Appen¬ 
dix IV. has been inserted. Formulae for moments and re¬ 
actions are there collected in the simplest form for applica¬ 
tion. 

W. H. B. 

Phcenixville, Pa., 

Feb. 24 th, 1886. 


CONTENTS. 


CHAPTER I. 

General Consideration of the Laws Governing the Action of Stresses in 

Trusses. 

PAGE 1 

Art. i. —The Truss Element. 1 

“ 2.—General Case. 3 

3. —Web and Chord Stresses in General. 4 

4. —Overhanging Truss—Parallel Chords—Bracing with two Inclina¬ 

tions . 8 

4< 5.—Overhanging Truss—Parallel Chords—Uniform Bracing—Vertical 

and Diagonal Bracing. II 

CHAPTER II. 

Special Non-continuous Trusses with Parallel Chords. 

Art. 6.—Distribution of Fixed and Moving Loads. 18 

7. —Position of Moving Load for Greatest Shear and Greatest Bending. 20 

8. —Fixed Weight. 36 

9. —Single System of Bracing with two Inclinations. 38 

“ 10.—Single System of Vertical and Diagonal Bracing—Verticals in 

Tension. 50 

“ 11.—Single System of Vertical and Diagonal Bracing—Verticals in 

Compression. 54 

“ 12.—Two Systems of Vertical and Diagonal Bracing—Verticals in 

Compression. 59 

“ 13.—Truss with Uniform Diagonal Bracing—Two Systems of Triangu¬ 
lations . 66 

“ 14.—Compound Triangular Truss. 69 

“ 15.—Methods of Obtaining Stresses—Stress Sheets. 74 

“ 16.—Ambiguity caused by Counterbraces. 74 


















Vlil 


CONTENTS , 


CHAPTER III. 

Non-continuous Trusses with Chords not Parallel. 

PAGE 

Art. 17.—General Methods. 75 

“ 18.—Curved Upper Chord—Two Systems of Vertical and Diagonal 

Bracing. 78 

“ 19.—General Considerations. 88 

“ 20.—Position of Moving Load for Greatest Stress in any Web Member. 89 

“ 21.—Position of Moving Load for Greatest Stresses in Chords.92/ 

“ 22.—Horizontal Component of Greatest Stress in any Web Member— 

Constant Value of the Same for Vertical and Diagonal Bracing 

with Parabolic Chord. 93 

“ 23.—Bowstring Truss—Diagonal Bracing—Example. 93 

“ 24.—Bowstring Truss—Vertical and Diagonal Bracing with Counters. . 102 

“ 25.—Bowstring Truss—Vertical and Diagonal Bracing without Coun¬ 
ters. 105 

26.—Deck Truss with Curved Lower Chord, Concave Downward—Ex¬ 
ample . io 3 

“ 27.—Crane Trusses. m 

“ 28.—Preliminary to the Treatment of Roof Trusses—Wind Pressure— 

Notation. 115 

“ 29.—First Example. 117 

“ 30.—Second Example. 118 

“ 31.—Third Example. 121 

“ 32.—Fourth Example... 122 

“ 33.—General Considerations.,... 123 

CHAPTER IV. 

Swing Bridges. Ends Simply Supported. 

Art. 34.—General Considerations. 125 

“ 35-—General Formulae for the Case of Ends Simply Supported—Two 
Points of Support at Pivot Pier—One Point of Support at Pivot 

Pier. 126 

“ 36.—Ends Simply Supported—Two Points of Support at Centre— 

Partial Continuity—Example. 133 

“ 37 *—Ends Simply Supported—Two Points of Support at Centre— 

Complete Continuity—Example. 159 

“ 37 a -—Tables and Diagrams—Turntables and Engines. 179 

“ 38.—Ends Simply Supported—One Support at Centre—Example .... 187 

CHAPTER V. 

Swing Bridges. Ends Latched to Supports. 

Art. 39.—General Considerations. 203 

4°.—Ends Latched Down—One Point of Support between Extremities 

of Truss—Example. 203 

























CONTENTS. 


IX 


CHAPTER VI. 

Swing Bridges. Ends Lifted. 

PAGE 

Art. 41.—General Considerations. 212 

“ 42.—Ends Lifted—One Point of Support between Extremities—Ex¬ 
ample. 212 

“ 43.—Final Observations on the preceding Methods. 222 

CHAPTER VII. 

Continuous Trusses other than Swing Bridges. 

Art. 44.—Formulae for Ordinary Cases—Reactions—Methods of Procedure. 224 

CHAPTER VIII. 

Arched Ribs. 

Art. 45.—Equilibrium Polygons. 229 

“ 46.—Bending Moments. 234 

“ 47.—General Formulae. 236 

“ 48.—Arched Rib with Ends Fixed. 237 

“ 49.—Arched Rib with Free Ends. 247 

“ 50.—Thermal Stresses in the Arched Rib with Ends Fixed. 250 

“ 51.—Thermal Stresses in the Arched Rib with Ends Free. 256 

“ 52.—Arched Rib with the Fixed Ends—/and n Variable. 259 

“ 53.—Determination of the Stresses in the Members of an Arched Rib—> 

Example—Fixed Ends—Consideration of Details. 267 

“ 54.—Arched Rib Free at Ends and Jointed at the Crown. 278 

CHAPTER IX. 

Suspension Bridges. 

Art. 55.—Curve of Cable for Uniform Load per Unit of Span—Suspension 
Rods Vertical—Heights of Towers Equal or Unequal—Gen¬ 
eralization . 279 

“ 56.—Parameter of Curve—Distance of Lowest Point of Cable from 

either Extremity of Span—Inclination of Cable at any Point... 281 

“ 57.—Resultant Tension at any Point of Cable. 282 

“ 58.—Length of Curve between Vertex and any Point whose Co-ordi¬ 

nates are x and /, or at which the Inclination to a Horizontal 

Line is i . 283 

“ 59.—Deflection of Cable for Change in Length, the Span remaining the 

same. 285 

'* 60.—Suspension Canti-Levers. 287 




















X 


CONTENTS. 


PAGE 

Art. 6i.—S uspension Bridge with Inclined Suspension Rods—Inclination of 
Cable to a Horizontal Line—Cable Tension—Direct Stress on 

Platform—Length of Cable. 288 

“ 62.—Suspension Rods; Lengths and Stresses. 292 

“ 63.—Pressure on the Tower—Stability of the Latter—Anchorage. 294 

“ 64.—Theory of the Stiffening Truss—Ends Anchored—Continuous 

Load—Single Weight. 296 

“ 65.—Theory of the Stiffening Truss—Ends Free—Continuous Load— 

Single Weight. 307 

“ 66.—Approximate Character of the Preceding Investigations—Deflec¬ 
tion of the Truss. 311 

CHAPTER X. 

Details of Construction. 

Art. 67.—Classes of Bridges—Forms of Compression Members—Chords 

Continuous or Non-Continuous.. 313 

“ 68.—Cumulative Stresses. 315 

“ 69.—Direct Stress Combined with Bending in Chords. 316 

“ 70.—Riveted Joints and Pressure on Rivets. 323 

“ 71.—Riveted Connections between Web Members and Chords. 325 

“ 72.—Floor-Beams and Stringers—Plate Girders. 326 

“ 73.—Eye-Bars or Links. 343 

“ 74.—Size of Pins. 345 

“ 75.—Camber. 350 

76-—Economic Depth of Trusses with Parallel Chords. 353 

“ 77 -—Fixed and Moving Loads . 354 

78.—Safety Factors and Working Stresses. 358 

79 -—General Observations. 365 

CHAPTER XI. 

Wind Stresses and Braced Piers. 

Art. 80.—Wind Pressure. 366 

“ 81.—Sway Bracing. 370 

82. —Transverse Bracing for Transferring Wind Stresses from One 

Chord to Another—Concentrated Reaction. 379 

83. —Transverse Bracing with Distributed Reactions. 384. 

“ 84.—Stresses in Braced Piers. 387 

85.—Complete Design of a Railway Bridge. 395 

APPENDIX I. 

The Theorem of Three Moments . 431 

APPENDIX II. 

The Resistance of Solid Metallic Rollers. 443 

APPENDIX III. 

The Schwedler Truss. 447 

APPENDIX IV. 

Reactions and Moments for Continuous Beams...453 

APPENDIX V. 


Cantilevers 


455 



































CONTENTS. 


X! 


APPENDIX V. 

Cantilevers. 

PAGE 

Art. i.—C antilever Structures—Positions of Loading for Greatest Stresses 

in the Cantilever Arm.455 

“ 2.—Positions of Moving Load for Greatest Stresses in the Anchor 

Arm, and the Greatest Stresses Themselves. 464 

“ 3.—Positions of Moving Load for Greatest Stresses in Anchor Spans 

and the Greatest Stresses Themselves. 467 

“ 4.—Wind Stresses. . 4^9 

* 5.—Economic Lengths of Spans and Arms. 47 2 

“ 6.—Wind Pressure... 474 









CHAPTER I. 



GENERAL CONSIDERATION OF THE LAWS GOVERNING THE 
ACTION OF STRESSES IN TRUSSES. 

Art. 1.—The Truss Element. 

A TRUSS may be defined to be a structure so composed of 
individual pieces that, if all the externally applied forces 
called loading are parallel in direction, the other external 
forces called reactions will be parallel both to each other and 
the loading. 

The simplest of all trusses is a triangle, and all trusses, 
however complicated, containing no superfluous members, 
are, and may be considered, assemblages of triangles simply. 
That the triangle is the truss element arises from the fact 
that it is the only geometrical figure whose form may not be 
changed without varying the lengths of its sides. 

In the elementary truss of the figure, let any force act ver¬ 
tically downwards at B, and consider the two triangles ABD 
and BDC having the common side BD. Since all external 
forces are parallel, the reaction at A is to the reaction at C as 
DC is to AD. 

For if BD be taken to represent the vertical force acting at 
B , and DF be drawn parallel to AB as well as EF parallel to 
AC, then will DF represent the stress in AB, and BF that in 
BC. But by the construction ED : EB — DC : AD, but ED is 
the vertical component of the stress in AB as well as the re¬ 
action at A, while BE is the same component of the stress in 
BC, and, similarly, the reaction at C. It is to be particularly 
noticed that EF is the common horizontal component in each 
of the members AB and BC, and also the resultant stress in 
AC. 


i 


i 


2 


THE ACTION OF STRESSES IN TRUSSES. 


When, therefore, the truss is horizontal, as is supposed in 
the figure, the vertical component in each of the members 
AB and BC is equal to the reaction at its foot; also the hori¬ 
zontal component of stress in each of these members is equal 
to the horizontal component in the other, as well as to the 
resultant stress in the third horizontal member. 

These simple principles constitute the foundation of all 
stress analyses in trusses. 


B 



Art. 2.—General Case. 

Again, consider any truss whatever, as that in Fig. I, in 
which the supports are not on the same level, nor are any two 
of the triangles of which it is composed similar. Suppose a 
vertical load to act at any apex, as A, the reactions will be 
vertical. Let the truss be cut by any plane which divides the 
line GH (AH is the trace of such a plane), then the part of 
the truss which is found on the left of AH is held in equilib¬ 
rium by a component of the vertical force at A, the vertical 
reaction at C, and the induced stress in GH. Since there is 
equilibrium, the lines of action of those forces must intersect 
in a point; and since the forces acting through C and GH 
have lines of action intersecting at D , the line of action of the 
component of the vertical force at A must pass through the 
same point. Thus the line of action DA for one component 
is established. 

In precisely the same manner BE is erected and produced 
until it intersects GH , produced, in E and the line of action 
AE of the other component established. Connect D and E } 
then, so far as the reactions are concerned, the case will not 





GENERAL CASE. 


3 


be changed if the actual truss be supposed displaced by the 
simple truss DEA. Let AN represent the vertical load at A, 
then make NO, parallel to AE, and DA, produced, intersect 
at O. If MO be drawn parallel to DE, AM will evidently 
represent the reaction at C, and MN the reaction at B. Pro¬ 



duce AN until it intersects GH in G, and draw DK and EF 
in a horizontal direction, then, from similar triangles: 

OM _DG 
MA ~ GA ’ 

Dividing one by the other, 

MN _ 

MA~ EG ~ EF * 

But DK 4 - EF is equal to the span, hence the reactions are 
inversely as the segments of the span, or any load or system 
of loading, vertical in direction, to which any truss whatever 
is subjected, is divided into reactions according to the law of 
the lever. Farther, whatever the internal stresses of the truss 
may be, at the ends the sum of the vertical components must 
be equal to the reactions. 


and 


OM EG 
WW~GA 


DG DK 


















4 


THE ACTION OF STRESSES IN TRUSSES. 


Art. 3.—Web and Chord Stresses in General. 

In the preceding case no account was taken of the stresses 
to which the individual members of the truss were subjected, 
but it will now be necessary to consider them. For this pur¬ 
pose FI. I., Fig. i, will be used, in which the points of support 
will be taken in the same level, the loading, vertical; and the 
truss will be considered as made up of similar triangles. The 
first and last suppositions in nowise affect the generality of 
the conclusions, but the operations are thereby simplified and 
given a character approaching more nearly to that of the 
ordinary operations of the engineer. 

In PL I., Fig. i, is the representation of a truss placed upon 
two supports A and L in the same horizontal line. CH and 
AL are parallel, and the oblique members included between 
them may have any inclinations whatever, only they are made 
symmetrical in reference to a vertical centre line through 0 . 
Let any weight W act at any point, as N, and let t and t' 
represent the tangents of the greater and less inclinations, 
respectively, of the oblique members to a vertical. Erect ver¬ 
ticals at A and L which will intersect the prolongations of 
CH in B and K\ then, as has been shown, BR and UK drawn 
through W will be the lines of action of the components of W 
which act on the two parts of the truss. The force parallelo¬ 
gram WRNU can then be drawn, in which RT and UV are to 
be drawn parallel to AL. NT — VIV is the reaction at A , 
and NV that at L. 

Resolve NR in the direction of AW and NM' by drawing 
M'R parallel to FN, then will M'R represent the stress in 
FN. The stress in FN will induce the stresses Fb and Fb", 
in FG and FO, and in the same manner the stresses shown in 
the figure will be induced at all the points on the left of FN. 

It is to be noticed that all the inclined stresses at the points 
A , C, Q , D, P, £, 0 , F, as well as M'R , have the same vertical 
component, NT ; also, that all the horizontal stresses at C, D y 
E, and F are equal to each other and act in the same direc¬ 
tion, each one having the value NT x (t + t’). Let n be the 


WEB AND CHORD STRESSES IN GENERAL . 


5 


number of the points C, D, E , and F, then the total horizon¬ 
tal stress acting along CH from left to right will be 


NT x (/ + {) x n. 


At N there is the horizontal force NM' acting from left to 
right. Let d equal the depth of the truss, or AB f and / and /' 
the segments AN and NL respectively of the span AL or s\ 
then from the figure it is seen that 



At the points 0, P, Q, A, there are horizontal stresses act¬ 
ing from right to left. From the diagram it is seen that the 
stress at O is (2 / x NT) ; at P, (t + t') x NT ; at Q, (t + f) x 
NT ; at A, /' x NT ; hence the total stress on the left of N 
acting from right to left is 



NT{nt -f (n — i)t') = NT 


Hence there is deduced the important result that NM' is 
just equal to the total horizontal stress on the left of N, and 
possesses the same line of action but is opposite in direction, 
therefore the two forces balance each other. 

Next prolong GN until it cuts UV in Y y then will NY 
represent the stress induced in GN by the component UN. 
The stress at N, acting from right to left, is therefore UY } or 



Although the diagrams are not drawn, it is plain that the 
horizontal stresses at M and L are NV x {t 4 - /') and NV x t’ 
respectively ; also, that their directions are from left to right; 
hence the total horizontal stresses, on the right of N\ which 
act from left to right, are 



6 


THE ACTION OF STRESSES IN TRUSSES. 


NVx{t + 2 /’) = Nv fffA. -^ =NVx (L- t y 

Hence the stress UY is balanced by the horizontal stresses on 
the right of N. All the internal horizontal stresses acting 
along AL are, therefore, balanced. 

According to the two force parallelograms drawn from G 
and H , it is seen that all the horizontal stresses on the right 
of N, which act from right to left along CH, are 

n' x NV x (/ + t') = NV x ^ , 

in which n is the number of apices G and H 

But NV — NTj ,; hence 

r l 

NV -j — NT -3. 
cl a 

But it has already been shown that all the stresses which 
act along CH from left to right are 

NT x (/ + f) xn = NT^. 

Hence all the horizontal internal stresses of the truss are 
perfectly balanced among themselves. 

This important characteristic belongs only to the “truss’" 
proper, and distinguishes it from all other bridge superstruc¬ 
tures. 

If an irregular truss, like that in Fig. I above, were treated, 
precisely the same result would be reached, but the resultant 
horizontal stress would be expressed by 2Pt or 2'Pt, in 
which P is a variable portion of W ; and d would be a “ mean ” 
depth, such that 21 d — /, and 12' td— 1. 

The portion FG is subjected to all the stress induced at the 
points C , D , E , and F ; EF to all that induced at C\ D, and 
E ; etc. It is important to notice this accumulation of stresses 
from panel to panel in the horizontal lines CH and AL. for 



WEB AND CHORD STRESSES IN DETAIL . 


7 


it shows that the stresses in those portions are not uniform 
from end to end. A stress induced at one point may be felt 
at any distance from that point. 

The upper and lower portions of the truss, CDEFGH and 
AQPONML , are called the top and bottom “chords,” and 
all members included between the chords, whether inclined 
or vertical, are called “ braces ” or web members. The various 
portions into which the chords are divided, usually equal to 
each other, are called panels. 

From the figure it is seen that the vertical components of 
the stresses in the braces or web members on one side of N 
are equal to each other; also, that the chord stresses have no 
vertical component, being horizontal. Farther, the vertical 
component in any brace or web member is equal to the reac¬ 
tion found on the same side of the load as itself. In a truss 
provided with horizontal chords, therefore, the office of the 
web members is solely to transfer, so to speak, the load from 
its point of application to the abutments or piers of the 
bridge; their duty, therefore, is precisely the same as that of 
the web in a flanged girder, hence their name “ web mem¬ 
bers.” In other words, the braces or web members take up 
the shearing stress at any section. 

Let sec i and sec i' be the secants of the angles of inclina¬ 
tion of the web members, corresponding to the tangents t 
and and let .S' and S' be the shearing stresses in the two 
segments of the span ; then the web stresses in the left-hand 
segment will be 5 sec i and 5 sec i', and those for the right- 
hand segment S’ sec i and 5 ' sec i 

The general principles brought out in the preceding results, 
therefore, are these: With horizontal chords the web stresses 
are products of the shears by the secants of the inclinations, and 
the chord stresses are functions of the tangents of the inclina¬ 
tions of the braces or web members from vertical lines . 

From an inspection of the force parallelogram at D , for 
instance, it is seen that the increment of chord stress at any 
panel point is equal to the algebraic sum of the horizontal com¬ 
ponents of the stresses in the web members intersecting at that 
point. The sum is numerical when, as in the figure, the 


8 


THE ACTION OF STRESSES IN TRUSSES. 


braces slope on different sides of the vertical line passing 
through the panel point, but the numerical difference is to be 
taken when both braces are found on the same side. 

Two web members intersecting at any panel point have 
stresses of opposite kind^ induced by the same shearing 
stress. 

The stress in CH is of course compressive, while that in 
AL is tensile. 

The preceding general results have been deduced on the 
supposition of the application of but one weight, but they 
are equally true for any system of loading. For the effect of 
any system of loading is simply the summation of the effects 
of the individual loads of which it is composed, hence only 
those principles which are true for the individuals can be true 
for the system, and those at least must hold, for the action of 
each load is independent of all the others. 


Art. 4.—Overhanging Truss.—Parallel Chords.—Bracing with Two In¬ 
clinations. 

Probably the simplest case of a truss subjected to the action 
of external loading occurring in the practice of the engineer 
is a simple truss fixed at one end, and is the case with one 
arm of a swing-bridge when open and subjected to its own 
weight as load. 

Now, in all cases of actual trusses, the load will be sup¬ 
posed divided in its application to the truss between the 
upper and lower chords. It will not, however, be equally 
divided, because the floor system of the bridge will rest 
wholly on one chord or the other. 



In the figure, let the truss be fixed at AB, and let W and 





OVERHANGING TRUSS. g 

IV be the panel loads on the upper and lower chords respec¬ 
tively, except at the extremity P of the upper chord, where 
\ W will rest. Let a represent the angle QPR, and a' the 
angle SQT ; the line PR is vertical, and AP horizontal. 

A simple and direct application of the principles and for¬ 
mulae of Art. 3 gives the following results : 


Stress in O . \W sec a. 


u 

“ M.... 

....(1 iW+ IV') 

u 

ll 

“ K .... 

....{ 2 IW+ 2 wj 

u 

u 

“ G .... 

....(3iW+ 3 W') 

u 

u 

“ E .... 

....(4\W+ 4 W') 

(( 

u 

“ D .... 

....(SiW+ slV) 

u 

u 

“ N .... 

....(iw+ W) 

sec a' 

(< 

“ L .... 

....(1 \W+ 2 W') 

u 

u 

“ H.... 

....(2iW+ 3W) 

u 

u 

“ F .... 

....(3iW+4 W') 

u 

u 

“ V.... 

....{ 4 \W+ 5 IV) 

u 

u 

“ c.... 

....(SilVp 6W) 

u 


It will be observed that in every instance the stress in any 
brace is the “ shearing stress ” in the section to which the 
brace belongs, multiplied by the secant of the inclination to 
a vertical line of the brace in question. For instance, if the 
brace L be divided by a vertical plane, the weights on the 
right of it, or the shearing stress, are (i \W + 2 W'), and this 
multiplied by the secant of a is the stress desired. 

The chord stresses are also determined by a direct and sim¬ 
ple application of the principles and formulae of the preceding 
article. 


Stress in f is \ W tan a 

“ “ e “ i \W tan a 4- ( W ' 4- \ W) ( tan a-\-tan «') 

“ “ d “ 2 \W tan a + (2 W+ 3 W') (tan a + ta?t ct r ') 

u u c « siWtan a + (4lW+6W) (tan «+tan a') 

“ “ b “ A^W tan a + (8 W-h 10PV') (tan ct + tan a’) 

“ “ a “ 5.1 Wtan a + (i 2 ^ W+ 15 W r ) (tan a + tan a), 



In the lower chord the stresses are as follows: 















10 


THE ACTION OF STRESSES IN TRUSSES. 


Stress in ^is 
“ “ h “ 

(( u u 

u u l u 

“ u m u 

a «< ^ u 


W' tan a' + iJV {tan a + tan a) 

2 W tan a + {2 W+ W') {tan a+tan a') 

3 W’tan a' + (4J W+^W') (tan a 4 -tan a ,s ) 
4 W' tan a 4- (8 W+6W') (tan a + tan a ! ) 

5 W’ tan a +{\2\W+ loW')(tan a + tana) 

6 W' tan a r + (i8JV+ I$IV') {tan a 4 - tan a') 


K 2 )- 


In determining these quantities, it is to be remembered that 
the stresses cumulate from the free end to the fixed ; i.e., the 
stress developed at any panel point is felt throughout those 
portions of the chords included between that point and the 
fixed end of the truss. 

General formulae for the Eqs. (1) and (2) may easily be 
found. Let n be the number of the panel, from the free 
end, in the chord AP {/is number 1 ; e, 2 ; d, 3 ; etc.), then 
the formula expressing the results in Eq. (1) is the following: 


Stress in any panel — {n — /) W tan a 4 - 



W + 



j tan a 4- tan a' | 



This expression gives the stress in any panel of AP. 

The formula which expresses the results shown in Eq. (2) 
is the following: 


Stress in any panel — nW' tan a' + | — IE 4 - 

n{n — 1) ytt f } c ^ 

---- W j j tan a 4- tan a | . . . (4). 

In which n denotes the number of the panel in the chord BQ 
starting from the free end ; i.e., g is number 1, h number 2, 
etc. 

The weight at P has been taken at half that applied at 
other panel points in the same chord. In the case of a swing- 
bridge, however, it is greater than that, since some of the 
details of the locking apparatus, etc., are hung from that 
point. Yet the Equations (1) to (4) may still be used, only a 









OVERHANGING TRUSS. 


II 


simple term is to be added to each of those equations. Let 
p be the panel length, d the depth of the truss, and W x the 
actual weight hung from P. Also, let W 1 — J W — w'. In 
order to find the additional stress produced in any panel d of 
the chord AP, let the moment of w be taken about the in¬ 
tersection of H and K in the lower chord ; this moment is 
w\(n — i)p 4- d tan a\. Consequently the additional stress 
desired is 


( / s p 

— w \ (n — i)—-\- tan a 
( d 



The stress s is to be added to each of equations (i) and (3) 
if W x > | W f otherwise it is to be subtracted. 

In precisely the same manner, the additional stress for the 
lower chord BQ is 

p 

s' = wn P . . ..(6). 

The stress s' is to be added to equations (2) and (4) if W x > 
\Wy otherwise it is to be subtracted. 

If, as in Fig. 1, AP is the upper chord, the stress in QP 
and all members parallel to it will be compressive ; while the 
stress in QS and all braces parallel to it will be tensile. Like¬ 
wise the stress in AP is tensile, and that in BQ compressive. 

If the truss were turned over so that BQ would become the 
top chord, the expressions for the stresses in equations (1) to 
(6) would remain exactly as they are, only the signs of the 
stresses would change. The condition of stress would be ex¬ 
actly represented in the preceding paragraph by simply chang¬ 
ing “ compressive ” to “ tensile,” and “ tensile ” to “ compres- 

• »y 

sive. 


Art. 5.—Overhanging Truss—Parallel Chords—Uniform Bracing—Ver¬ 
tical and Diagonal Bracing. 

The two most frequent cases of Fig. 1, Art. 4, are, first, 
that in which a = a\ and, second, that in which a — o. The 
first of these cases is represented in Fig. 1, and the second in 
Fig. 2. 



12 THE ACTION OF STRESSES IN 7RUSSES. 

The web stresses for this case will be precisely the same in 
general form as those given in Art. 4, but sec a will be written 
for sec a\ 

Very simple general formulae can be written for these web 
stresses. Let ri denote the number of any brace starting 
from 0, which is called 1 ; then observing the general values 
in Art. 4, the stress in any brace n' parallel to 0 will be 

4- b — | — W + ^ U - -—— W | sec a 4 - w sec a. . . (1). 



The expression ( 4 - b), of course, denotes tensile stress in any 
brace parallel to 0. 

In precisely the same manner, the compressive stress in any 
brace parallel to N (ri possessing the same signification as 
before ; i. e ., ri for TV is 2 ; for A, 4, etc.) is 

-b- | ~ ^ W + ~ W l sec a + w sec a . . (2). 

In determining the chord stresses, it is to be remembered 
that the weights W rest on the lower chord AP. Making 
tan a = tan a in Eq. (3) of Art. 4, the stress in any lower- 
chord panel is 

C — (n — J) Wtan a 4 - | (n— 1 ) 2 W+n (n— 1) W" j tan a + s . (3). 

Making the same change in Eq. (4) of the previous Article, 
the upper-chord tensile stresses will be found to be 

nW'tan \ n 2 W+n (n-i)W'\ tan a-\-s' . . . (4). 







OVERHANGING TRUSS. 


13 

Some of the results given by the formulae should always be 
checked by the method of moments. 

Let it be desired to determine the stress in k by the method 
of moments. Let the origin of moments be taken at the 
intersection of G and H. The moments which balance each 
other about that point are that of the stress in k acting with 
the lever-arm d, the depth of the truss, and those of the 
weights applied to the truss on the right of the panel point 
in question ; these latter act against the former. Calling the 
panel length and taking the moments mentioned: 

Td = 3W' • + 2W • \p + • 3/ + w • 3/. 

••• T=4kW’£ + 4iW^+ 3w£ ' • • (5) ' 

The result of Eq. (5) ought to be the same as that of Eq. 
(4). Two or three panels in each chord ought to be treated 
in the same manner. 



The cantilever truss represented in Fig. 2 shows the rase 
in which a' of Fig. 1, Art. 4, is equal to zero. The notation 
is precisely the same as that used before. 

The general expression for the stress in any inclined brace 
is simply Eq. (1) repeated—that is: 

+ b = | W + —— W f | sec a + w sec a . . (6). 

Making sec a — 1, there results for the compressive stress 
in any of the verticals 2, 4, 6, etc.: 











14 


THE ACTION OF STRESSES IN TRUSSES. 


- 0 = j — 2 — W + y W' | + w . (7). 

Making tan a' = o, in Eq. (3) of Art. 4, gives the com¬ 
pressive chord stress in any panel of the lower chord AP. 
Hence, for that chord: 

C = | (» - *) W + ^ W' | tana + 

wn I .. . ( 8 ). 

a 

In a similar manner, from Eq. (4) of the previous article, 
for the tensile stress in any panel of the upper chord, there 
results the equation* 

_ j 1? 1Tr n (n — 1) iTTj ) P /x 

T — j — fE + — W j- tan a 4- wn ^ . . . (9). 

If W x should be less than J IV, the term which expresses 
the additional stress, whether in braces or chords, will be sub¬ 
tractive, as will be indicated by the sign of w. 



Again, applying the moment test to any panel, as c , by 
taking the origin of moments at R, the notation remaining the 
same as before, there results: 

Cd — 3 (W 4- IV) . 2p 4- J W . 4p + w . 4/> 

C = (4W + + 4 wt . . . (,o). 


















OVERHANGING TRUSS. 


15 

This result ought to agree with that shown by Equation (8), 
and several panels in each chord should be tested. 

In the great majority of cases it is not convenient to apply 
a general formula, but the numerical values are usually deter¬ 
mined directly from the diagram, and the stress in each mem¬ 
ber written along it, as shown in Fig. 3. 

Fig. 3 shows a truss which frequently occurs in the practice 
of the American engineer; it is in reality one arm of an open 
swing-bridge. 

Let the panel length — p — 12 ft., and the depth of the 
truss — d — 20 ft. The tangent of CQP — 12 -f- 20 = 0.6, and 
the secant of CQP — 1.166. The tangent of DEP — 1.2, and 
the secant of DEP — 1.562. The panel loads at E, E f //, etc. 
= W' — 3.00 tons; at C , D , G, K, etc., W— 5.00 tons, at 
P, W x — 4.00 tons. No load is taken at Q. 

In the figure there are two systems of right-angled triangu¬ 
lation ; P, E , D, H , K , etc., is one system, and C, E, G, L , M, 
etc., is the other. This does not, however, complicate the 
matter in the least, for each system of triangulation is regarded 
as an individual truss carrying its own weights only. Calcula¬ 
tions are therefore made for each system of triangulation as 
if they were independent trusses, and then the two are 
added. 

The weight of the portion EQP is supposed divided be¬ 
tween E and P, thus showing W x > }W. The vertical braces 
are evidently in compression, while the inclined ones are in 
tension. 

The figures in the diagram denote tons (2000 lbs.) of stress; 
+ indicates tension, while — indicates compression. 

Stress in PE = W t sec DEP — 4 x 1.562 = 6.248 tons. 

“ “ CF — W sec DEP— 5 x 1.562 = 7.810 “ 

“ “ DH=(W 1 4- W’ + W)x 1.562 = 18.744 “ 

« “ GL— 2W+W'x 1.562 = 20.306 “ 


The other brace or web stresses are found in precisely the 


same manner. 


16 THE ACTION OF STRESSES IN TRUSSES. 

Stress in CP — W 1 tan DEP —4 x 1.2 

= 4.8 tons, 

“ “ DC — W tan DEP + 4.8 = 6.0 4- 4.8 

= 10.8 tons. 

“ “ GD— (W t + W' + W) x 1.2 + 10.8 = 14.4 + 10.8 

= 25.2 tons. 

“ “ KG = (IV + W' + W) x 1.2 + 25.2 = 40.8 tons. 

Other lower-chord stresses are found in exactly the same 
manner. 

By an inspection of the diagram it is seen that the general 
expression for the stress in EE is precisely the same as that 
for CP; the same can be said of HE in reference to DC ; LH 
in reference to DG ; etc. The explanation of this is simple. 
If the truss be divided by a plane normal to the paper and 
parallel to the inclined braces, only vertical and horizontal 
members will be cut. But the truss is in equilibrium, and 
since the loading is wholly vertical, the sum of the horizontal 
stresses must be zero; or, the stress in the lower-chord panel 
cut must be equal and opposite to the stress in the upper- 
chord panel cut. 

Let the moment test be applied to the stress in the panel 
MK. The origin of moments for the loads applied to the 
system PEDH , etc., is N, and the origin for the other system 
is L. Taking moments about those points : 

C’d + C"d = 16 x 36 + 4 x 72 + 8 x 24 4- 5 X 48 
.'. C — C' + C" = 64.8 tons. 

Again, for the lower-chord panel adjacent to A, B is the 
moment origin for the whole load. 

C — (48 x 42 + 5 x 84 + 4 x 96) 7- 20 = 141 tons. 

Thus the numerical results are verified. 

According to one of the principles of Art. 3, the horizontal 
component of the stress in any inclined web member ought 
to be equal to the increment of chord stress at either of its 


OVERHANGING TRUSS. 


17 


extremities, and such will be found to be the case. If, for 
example, 20.3 be multiplied by the cosine of the angle GLH ’ 
the result will be 15.6 = 40.8 — 25.2. 

This last is a verification of the web stresses, and both 
methods of checking are perfectly general and may be applied 
to all trusses, as should be done in actual cases. 


CHAPTER II. 


SPECIAL NON-CONTINUOUS TRUSSES WITH PARALLEL 

CHORDS. 

Art. 6.—Distribution of Fixed and Moving Loads. 

The trusses treated heretofore have been of rather an ele¬ 
mentary character, and general principles have been con¬ 
sidered instead of special and practical applications. Before 
taking up the technical treatment of trusses it will be neces¬ 
sary to consider some preliminary matters. 

The total load on a bridge-truss always consists of two 
parts, the fixed load and the moving load. The fixed load 
consists of the entire weight of the bridge, including tracks, 
flooring, etc. The moving load, as its name indicates, con¬ 
sists of that load (whether single or continuous) which moves 
over the bridge. 

The truss is, of course, always subjected to the action of 
the fixed load. 

If the truss is of uniform depth the panel-fixed loads will 
be uniform in amount for one chord ; but if the depth is vari¬ 
able it may be necessary to make a varying distribution of 
the weight of the trusses and lateral bracing. The amount 
and rate of this variation can only be determined by the cir¬ 
cumstances of each particular case. 

The moving load on a railway bridge may be taken as con¬ 
tinuous or as a series of single weights as actually applied at 
the wheels under the locomotives and cars. The assumption 
of continuity of moving load was formerly always made, a 
larger amount per lineal foot being taken to represent the 
extra locomotive weight. In such a case, if the moving load 
extends from the end of the bridge to the centre of any 
panel, or to the end of that panel, the panel point immedi- 

18 


RAIL WA Y BRIDGES. 


l 9 


ately in front of the train will not sustain a full panel load; 
but if it be assumed that this panel point does sustain the 
full load, then a small error on the side of safety will be 
committed. Such an assumption was formerly made, and 
the consequent method of computation will be given in some 
of the Arts, which follow in this chapter. 

At the present time (1885), however, the demands of the 
best practice require the moving load to be taken at the 
actual points of application of locomotive and car wheels. 
This method of computation will be given in several of the 
first cases taken. 

If the span is short, or less than 125 feet, the moving 
load should be taken entirely of locomotives, as two will 
nearly cover the structure. The amount and character of 
the moving load, however, is usually indicated by specifica¬ 
tions. 

The moving load of a bridge may pass along the upper 
chord or the lower chord. In the first case the bridge is 
called a “ deck" bridge, and in the latter case a “ through ” 
bridge. The methods employed in the determination of 
stresses in the various truss members are exactly the same in 
both cases. 

“ Pony ” trusses are through trusses not sufficiently high or 
deep to need overhead cross-bracing. 

Every truss-bridge is composed of the following parts: 

Upper and lower chords, 

Upper sway-bracing, 

Web members, 

Floor system, including beams, stringers, ties, floor-hangers, 
lower sway-bracing, and rails. 

The sum of the weights of the parts is the “fixed ” load of 
the bridge. 

In the case of highway bridges the calculations are pre¬ 
cisely the same as for railway bridges, except that the mov¬ 
ing load is assumed to be uniform per lineal foot of bridge. 
The greatest load that can ordinarily pass on a bridge is a 
dense crowd of people, the greatest weight of which can be 
taken at eighty-five pounds per square foot. The late Mr. 


20 


SPECIAL NON-CON TIN U0 US TRUSSES. 


Hatfield, of New York City, found by experiment that it was 
scarcely possible to exceed seventy pounds per square foot. 
The moving panel load of a highway bridge may then be 
found by multiplying the width of the clear way, including 
sidewalks, by the product of the panel length with the load 
per square foot. 

If the span is not over 125 feet, or about that value, the 
moving load for the truss members may be taken at eighty- 
five pounds per square foot, or sixty pounds for greater 
lengths. In all cases, however, the floor beams and joists 
should be designed for a moving load of 100 pounds per 
square foot, in order to provide for the increased fatigue of 
those members due to shocks and sudden application of 
loads. 

In some cases highway bridges are subjected to enormous 
concentrated loads of a special character. Such loads can 
only be known from local considerations, and the bridges 
must be built with a view to sustaining such special weights. 

All the methods or principles used, then, in the following 
cases, which will be those of railway bridges, are equally ap¬ 
plicable to highway structures, and no further special atten¬ 
tion will be given to the latter. 

Art. 7.—Position of Moving Load for Greatest Shear and Greatest 

Bending. 

That method of computation which treats the moving load 
as composed of a system of isolated weights requires some 
simple method of finding the greatest possible shear in a 
given panel for a given system of loading. Among the first 
to use such a method was Theodore Cooper, C. E. ; and the 
results of the following investigation are the same as those 
determined by him. 

The method and formula first developed apply to any 
single system of triangulation so far as the web stresses are 
concerned, but for the chord stresses they only apply to such 
a system when composed of alternately vertical and inclined 
members. Subsequent modifications for web members all 
inclined will be made for the chord stresses. 


GREATEST STRESSES. 


21 


Case I. 

Let the moving load consist of the advancing weights W u 
W 2 , W 3 , .... W n separated by the distances a, b , 

c , d , etc.; then let the weights W±, W 2 , .... Wn' be 



found in the panel DC, in which it is desired to find the 
greatest possible shear, and at the distances b', b”, . . 

from C; it being understood that the moving load advances 
from F toward G. The last load W n is found at the distance 
x from F. The length of span FG is /; while the length of 
panel DC is p. With the assumed position of loading, the 
reaction R at G will be: 


R = 


W t 


ci b -\- c . T x 


+ W 2 


+ W 3 


b c -\- d . . . . . x 


l 


c + d + 


-p x • • • (i) 


+ w n x 7 


The parts of the weights W lf W 2 , 


. . resting on 















22 


SPECIAL NON-CONTINUOUS TRUSSES. 


b' b" 

DC, which pass to D, are W x —, —, . . . . Hence 

P P 

the shear in the panel DC will be 

S = R-(wP + w£+ .) .... (2) 

If the train advance by the amount A x, the new reaction 
R', at G , takes the value: 

R' — R + ( W\ + W% + +••••+ Wn) —jr * • ( 3 ) 

and the new shear will become : 

S'=R'-(W^+wP + . . . )-(lV t +IV 1+ ...)—. (4) 

.-. S' = S+(IV 1 + W, + IV S + . . . + W n ) ~ 


-(Wi+ W 2 + . . . )y-.(5) 

Or, 5 ' - S = ~- | (lV t + W, + W s + . . . + W n ) 

~(W 1+ W 2 . . . )(£=*)[.( 6 ) 


Whenever S' — S becomes equal to zero, 5 'will be either a 
maximum or a minimum. If the difference is positive just 
before it becomes equal to zero, S' will be a maximum, and 
that is the only case of interest in the present connection. 
Hence, by placing S' — S equal to zero, the following condi¬ 
tion is obtained : 

n(W, + W 2 + ...)= W, + W 2 + W 3 + . . . + W n . . (7) 

The shear in the panel in question will therefore take its 
greatest value when n times the moving load which it contains 
is equal to, or most nearly equal to, the entire moving load on 
the bridge . 









GREATEST STRESSES. 


23 


That equality will seldom or never exist unless one of the 
weights W is placed on a panel point, since W x + W 2 + . . . 
is seldom or never an exact divisor of the entire load on the 
bridge. If a weight rests on a panel point, any part of such 
a weight may be taken as acting in one adjacent panel and the 
remainder in the other ; the desired equality may thus be 
obtained. 

In case Eq. (7) should hold, the position of the moving 
load is a matter of indifference so long as the panel in ques¬ 
tion contains the same number of loads JV t + W 2 + . . 

as there is no trace of b' , b" , etc., in that equation. A load 
may then always be taken as resting at the rear extremity of 
the panel where the greatest shear in it exists. 

These considerations frequently essentially simplify com¬ 
putations. 

When the value of x has been found for the position of 
the greatest shear, the latter being determined by the preced¬ 
ing method, Eq. (2) may be put in the following convenient 
shape by the aid of Eq. (1): 

S — -j [ W\ ci + ( W t + W < 2 ) b 4- (Wx + W 2 4- W s ) £+•••• 

4 - 

+ (W t + W,+ . . . W n ) x] - j [Wi a + (W t + W,) & 

+ . . . +{W 1 + IV 2 + . . . +W r „._ I )?] • (8) 

The sign (?) stands for the distance between the wheel con¬ 
centrations W n >_! and W n <, since the latter rests directly at 
the panel point in question. 

It is thus seen that all the parts of Eq. (8) may be taken at 
once from tables, except that term involving 

Case II. 

In the preceding case it has been supposed that for the 
greatest shear in DC , the front weight W x is found between 
D and C; but let W l +W 2 + etc., be supposed between D 

and G. 


24 


SPECIAL NON-CON TIN UO US TRUSSES. 


With the notation remaining the same as before, the shear 
5 will become: 


b' b" 

S = R-{W l + W t + etc.) -{Wl + Wij + 


) I ( 9 )- 


while S' takes the value: 

S' = S + ( Wx T W 2 + + . . • + J^n) l 

-(Ws + w,+ . . . )^. 

Hence for a maximum, the following expression must never 
become negative: 

S’ - S = ~ \ (W, + W 2 + W 3 + . . . +w n ) 

— (+ . . . ) = ft'j | = o. * . (io). 

But Eq. (io) is identical with Eq. (7). Hence, the same 
conditions for a maximum obtain wherever may be the head 
of the moving load. The second member of Eq. (8), how¬ 
ever, must contain the negative sum of all the weights be¬ 
tween D and G. 


EXAMPLE. 

If each one of the weights W lt W 2 , etc., is equal to any 
other, i. e. , if they are all uniform, and if a = b — c = d 
— . . . = /, Eq. (7) shows that the front weight W x 

must be taken at the first extremity of the panel in question. 
The same result holds if the first weight W x is not exceeded 
in amount by any that follows it, provided that a , b , c , etc., 
still equal p. 

In cases where the same system of concentrated loads is to 
be used for a number of spans, it will be shown that the 
tabulation of the products of the sums of the weights W x , W 2 , 
etc., by the distances a , b , c, etc., can be advantageously used 




GREA TES T S TRESSES. 


25 


to shorten and simplify computation. In other cases, how¬ 
ever, the quickest and simplest method is partly graphical; 
it is as follows: 


C 



In Fig. 2 let AB be the length of span, and W any weight 
resting anywhere in the span. Erect a vertical at B and let 
BC represent W by any convenient scale; then draw the 
straight line A C. The vertical intercept WD will represent 
the reaction at B due to W, by the 
same scale on which BC represents 
that weight. 

In the same manner if BF repre¬ 
sents W 1 , then W X E will represent 
the reaction at B due to W 1 . Thus 
there must be as many verticals qJt 
BF , BC ', etc., as there are different 
weights resting in the span, and the 

total reaction at B , for any given position of the moving 
load will be the sum of the vertical intercepts WD, W 1 E , 
etc., erected at each load W for that position. 

b' b" 

The negative shears W\ -, W2 —, etc., appearing in Eq. (2) 

are most readily found in the same manner. If GH , Fig. 3, 
is a panel length and W any weight represented by the verti¬ 
cal line GK, drawn to any convenient scale, while WH is 
equal to b 1 , b n , etc., then the vertical intercept WE between 

b' b" 

GH and KH will represent hF—, W-, etc., i.e., the reaction 



, W- 
P 7 / 


at G due to W. 










2 6 


SPECIAL NON- CO N TIN UO US TRUSSES. 


If the reaction at B , Fig. 2, is then given by 2 WD , and 

JF,- + 0 'i- + etc. = Fig. 3, the shear (see Eq. (2)) 

/ P 

will be : 

S= 2 WD- 2 WE. 


If Figs. 2 and 3 are drawn on profile, or cross-section paper, 
the shears for any span can be found with great ease and 
rapidity. 


Position of Moving Load for Greatest Bending Moment. 


The Fig. and notation of the preceding cases will be used 
in connection with this. Moments will be taken about the 
panel point C, horizontally distant l from G. There will 
be supposed to be n! weights in front of C (i. e ., between 
C and G), and the weight W n > will be taken at the distance 
x from C towards G. The bending moment M will then 
take the value: 


M = Rl’ 


JVi ft -f- b -4- c *4“ 

T VP 2 ( b + c 4- 

- 

. . . -f- x ) 

. . . fx) 

. + W n ,x\ 



Or, after taking the value of R from the preceding cases: 

M=fw l a + (W x + W 2 ) b + {W, + W, + W,)c + . . . 

+ (fF"i + JV 2 + JV 3 + . . . + Wn) x ~\ — Wia — ( IV I + IF 2 ) b 
-(W 1 + W 2 + W 3 )c- . . . + IV 2 + W 3 + . . . 

+ VVf)x' ..( 11 ). 

If the train advances by the amount Ax, the moment be¬ 
comes : 

M' = M + flV 1 + W a + W,+ . . . +W n ) ax-(W 1 + W, 

+ • • • + W n >) &X, .(l2). 






GREATEST STRESSES. 


27 


Hence, for a maximum, the following value must never be 
negative: 

M — M — Ex | ~j ( W\ 4 - W 2 4 - IV$ 4 - . . . 4 - W — (JVi + W 2 

+ • • . +W r n')\=0 .(13). 

Or, the desired condition for a maximum takes the form: 

R _ Wi 4 - IV 2 4 - . . . 4 - W n < , 

1 - . . . + W n ■ ■ U4) - 

It will seldom or never occur that this ratio will exactly 
exist if W n < is supposed to be a zvhole weight ; hence, W n < will 
usually be that part of a whole weight at C which is neces¬ 
sary to be taken in order that the equality (14) may hold. 

It is to be observed that if the moving load is very irregu¬ 
lar, so that there is great and arbitrary diversity among the 
weights W, there may be a number of positions of the train 
which will fulfil Eq. (14), some one of which will give a value 
greater than any other ; this is the absolute maximum desired. 

Since W n > will always rest at a panel point for the greatest 
bending moment, x' in Eq. (11) may always be put equal to 
zero when that equation expresses the greatest value of the 
moment. The latter then becomes: 

M= j [W x a + (Wt + W,) b + . . . + ( W l +W,+ . . . 


+ W n )x]-W 1 a- (\\\ + Wi)b- . . . -(IVt+W, 
+ . . . +W n ._,)?.(15). 


In this equation, of course ;r corresponds to the position of 
maximum bending, while the sign (?) represents the distance 
between the wheel concentrations W n <_ x and W n >. 

It is known that for any given condition of loading the 
greatest bending moment in the beam or truss, will occur at 
that section for which the shear is zero. But if the shear is 
zero at that section, the reaction R must be equal to the sum 
of the weights (W x + IV 2 + + . . . + fE n .) between G 

and C; the latter now being that section at which the greatest 





28 


SPECIAL NON-CONTINUOUS TRUSSES. 


moment in the span exists. Hence for that section Eq. (14)* 
will take the form : 

V R 

7~ W 1 + W, + W 3 + . . . + wj 

or, the centre of gravity of the load is at the same distance 
from one end of the truss as the section or point of greatest 
bending is from the other. In other words, the distance between 
the point of greatest bending for any given system of loading , 
and the centre of gravity of that loading is bisected by the cen¬ 
tre of span. 

If the load is uniform, therefore, it must cover the whole 
span. 

It will be observed that Eq. (15) is composed of the sums 
of W lf 4 - W 2 , etc., multiplied by the distances a, b, c, etc., 
precisely as in Eq. (8), hence the same tabulation as there 
indicated may be used to advantage. 

Limitations of the preceding methods. 

The preceding methods are limited to a single system of 
triangulation. By the use of certain assumptions in refer¬ 
ence to the distribution of the loading between two or more 
systems of triangulations in the same truss, a somewhat simi¬ 
lar investigation might be made for such cases, but such 
analysis would not be rigorously exact. Hence it is as well 
to pursue the usual method and assume that each system 
acts as an independent truss, then place the moving load in 
such a position for each system that the front panel load for 
that system will be the greatest possible. This panel con¬ 
centration will then be the forward panel-moving load, and 
the succeeding ones may either be those concentrations 
which actually correspond to the forward one, or may be 
supposed to be composed of a uniform load equivalent to 
the concentrated one. The latter plan will be employed 
hereafter. 

Application of the preceding method to an all-inclined web 
system. 

As was observed at the beginning of this Art. the analy¬ 
sis for chord stresses, as already given, is directly applicable 



GREATEST STRESSES. 


2 9 


to a single system of triangulation in which a vertical web 
member is found in each panel. The general demonstration, 
however, is easy. 

If the web members are all inclined, the formulae, as al¬ 
ready given, are directly applicable in any case to the deter* 
mination of stress in that chord which does 7 iot carry the 
moving load, since moments are taken about the panel points 
of the chord traversed by the moving load. 

But let it be required to determine the position of the 
moving load for the maximum stress in DC, Fig. 1, and the 
expression for the corresponding moment. As before, let 
the load move from A toward G. Let q represent the hori¬ 
zontal distance of D from H, i. e., q — KH; evidently q is 
constant for the same span. Let x x represent the horizontal 
distance from H of the first load to the left of D , and let that 
load be represented by W\. That portion of the loads rest- 

ing in DC, which is transferred to D, is 2 W —. /' will now 

represent GD 4- q . 

By taking moments about H, Eq. (11) will take the form: 

Ml — — [ Wi a 4- ( W\ + IV 2 ) b 4 - ... + ( IV i + 4- . . . 4 - IVn) 

— W x a — ( IV 1 + fF>) b — ... — ( lV t + W 2 4- . . . + W 1 n ) x\ 


- < L 2 Wb' .(16). 

P 

By advancing the train Ax, since Ax = Ax t = Ab\ Eq. 
(13) will become: 


M —M— Ax ^ (Wi+ W 2 4- . . . 4- IV n ) — ( IV 1 4- IV 2 + . . .4- IV n ) 

. - q -2W | — o.(17). 

P ) 

The condition for a maximum or minimum then takes the 
shape: 

r _ w x +w % 4- .. . + w n 4- j 2 iv 

l IV 1 4- W% 4- W$ 4- . . • . 4- IV n 


. (18). 








30 


SPECIAL NON-C0NTINU0US TPUSSES. 


Eqs. (16) and (18) are the general expressions of which 
Eqs. (15) and (14) are special forms. 

After x and, hence, x x have been determined by the aid of 
Eq. (18), the maximum moment will be given by Eq. (16), in 
which the tabulations already indicated can be advantage¬ 
ously employed. 

Application of preceding methods to a system of concentra¬ 
tions followed by a uniform load . 

If the uniform load does not reach to the panel under con¬ 
sideration, which is usually the case, W n in Eqs. (7), (14) and 
(18) represents the total uniform load on the bridge, but the 
formulae are in no wise changed. In Eqs. (8), (15) and (16), 
however, it is to be observed that while W n again represents 
the total uniform load, x will represent the distance from its 
centre of gravity to the end of the span (i. e ., half the length 
covered by the uniform load), also that the distance between 
W n and W n _i will be equal to ;r plus the space which sepa¬ 
rates Wn_i from the front of the uniform load. 

In the case of the existence of this uniform load it will 
happen that W n < will not rest at a panel point. The last term 
in the negative expressions of the second members of Eqs. 
(8) and (15) will then be {W 1 + W 3 4- . . . + W n ) x ; 

x 1 being the distance of W n < in front of the panel point C. 
Eq. (16) is general, and needs no change on this account. 

If the concentrations are so few that the uniform load ex¬ 
tends over a portion of the panel in question, the observa¬ 
tions made above still hold. But in addition to them, W n < or 
W'n will represent the amount of uniform load in the panel, 
and x' or x x will represent the distance from its centre of 
gravity to the panel point. The interval or space between 
W n '_ i or W 1 ^ and W n < or W^ n will then be the distance from 
either of the former to the centre of gravity of the uniform 
load. 

Finally, in Eq. (17) or (18) 2 W will be either wholly or 
partly composed of uniform load. 

Modifications for Skew Spans. 

If a skew bridge is under treatment, the preceding methods 


GREA TEST STRESSES. 


31 


apply in all respects, so far as the general principles are con¬ 
cerned. 

It will be sufficiently accurate in all cases to treat the 




Fig. 4. 



moving load as if it were passing along the centre line LL 
LL' of each track. 

In the case of the single track bridge, Fig. 4, if the load is 
passing from right to left, the moving load does not rest on 
the truss CD until it passes the point C\ and continues to act 
on that truss until it passes to the same distance to the left 
of D, it being borne in mind that all moving load is trans¬ 
ferred to the trusses by transverse floor beams placed nor¬ 
mal to the axis of the bridge. It results from these con¬ 
siderations that if the load passes from right to left in Fig. 
4 and along LL , the reactions at D will be greater than 
a half of those at K by the amount of the half products 















32 SPECIAL NON-CON TIN UO US TRUSSES. 

of the total load corresponding to those reactions by the 
ratio 

CC' 

CD 


Hence, if R is any reaction at K and 2 W the total load, 
the corresponding reaction at D will be: 


R CC 
2 2 CD 


2 IV. 


On the other hand, with the load moving in the same 
direction, the reaction at A will be : 


4 


R_.CC' 

2 * CD 


2 IV. 


In Eqs. (8), (15), and (16), then, there is to be written for 
( IV. 4 - W 2 4 * . • . + I Vn) x 

the expression, 

(W. + W 2 + . . . 4 - W n ) (x ± CC% 

according to the direction in which the train is moving; but 
the negative poj'tions are to remain unchanged. It is to be 
remembered that the quantity x is to be measured on the 
centre line. 

In the case of the double track skew bridge of Fig. 5, in 
which there are the two trusses AB and CD only, precisely 
the same observations hold. For one track, however, CC" is 
to be used and CO for the other. Separate computations 
are to be made for each track for each truss. 

If there are three trusses in Fig. 5, each pair of trusses con¬ 
stitutes a single track bridge for the track between, and is to 
be treated precisely as Fig. 4. 

If the skew is so great that one or more floor beams have 
their end or ends resting on the masonry, obvious modifica- 




GREATEST STRESSES. 


33 


tions must be made according to the preceding general prin¬ 
ciples. 

The Graphical Method. 

With convenient means for constructing an accurate equi¬ 
librium polygon with a large number of loads, this method is 
a very rapid one for either shears or moments. A perfect 
familiarity with the principles and operations of Art. 45 is 
here supposed. 

Let the entire moving load for a given truss be represented 
by the system of forces 1, 2, 3, 4, and 5, in Fig. 6. They are 
given in actual position under the polygon PKO. P is the 



pole, EF the load line, and PO an indefinite horizontal line 
normal to EF. As usual, the moving load is supposed to 
move from O toward P. The polygon PQK .... NO is 
formed in the ordinary manner by making its sides parallel 
to the lines radiating from P. For reasons that will pres¬ 
ently be evident, QF should be continued considerably 
beyond C, while NO should be carried somewhat below 
the horizontal line through P. 

After the constructions indicated have been made to any 


I 


3 













34 


SPECIAL NON-CONTINUOUS TRUSSES. 


convenient scale, let the span under consideration be laid 
off to the same scale by which the horizontal separations 
between the loads, I, 2, 3 and 4, etc., are laid down, and let 
indefinite vertical lines be drawn through the panel points, 
but let all this latter construction be made on tracing cloth. 
In order to avoid confusion, neither the panel points nor the 
vertical lines through them will be shown in the Fig. 

In the first place, let the position of the moving load for 
the maximum shear in some web member be determined by 
Eq. (7), and in this position let the loads, 1, 2, 3 and 4, be 
supposed to rest on the truss ; and let B\ represent the dis¬ 
tance between the last concentration and the right hand of 
the span (i. e., B\ is x of Eq. (8) ). Now let the tracing 
cloth be superimposed on the equilibrium polygon in such a 
manner that AB shall represent the span ; then erect the 
verticals BD and AC, and draw CD, to the latter of which 
PH is drawn parallel. FH will then represent the reaction 
at the left end of the span, i. e., at A. If from this reaction 
the negative shear shown in Eq. (8), or found by the method 
of Fig. 3 be subtracted, the result will be the shear desired 
in the web member under consideration. In this manner all 
the maximum shears may be found. 

Again, let it be required to find the greatest moment at a 
given panel point, for which the moving load has been found 
by Eqs. (14) or (18), to occupy such a position that the dis¬ 
tance from the right end of the span to the last concentra¬ 
tion ( 1 . e., x in Eqs. (15) and (16) ) is represented by B' 5 in 
Fig. 6. Also with the position of moving load thus deter¬ 
mined, let it be supposed that the load 2 rests at the panel 
point considered. Now, let the tracing cloth be so super¬ 
imposed on the equilibrium polygon that A'B' will represent 
the span, then erect the vertical lines B'D' and A'C , and join 
C'D'. Since load 2 was found at the panel point, at which 
the moment is to be determined, KL will represent the max¬ 
imum moment in question. Each linear unit in KL, meas¬ 
ured by the same scale to which the span and distances, 1-2, 
2-3, etc., are laid down, will represent as many moment units 
as there are force units in the pole distance FT. If FT is 


GREATEST STRESSES. 


35 


in pounds and KL in feet, the PT X KL will be the mo¬ 
ment desired in foot-pounds. In the same manner all maxi¬ 
mum moments may be determined. In every position of 
moving load required, the vertical intercept between the 
closing line and equilibrium polygon, drawn through the 
panel point considered, will represent the maximum moment 
at that point. PH' drawn parallel to C'D' will give FH' as 
the reaction at P, though it is of no special value in this con¬ 
nection. The application of this method to the different 
panels of a truss will give all the greatest chord stresses. 

This method is, of course, subject to all the modifications 
that have been outlined for the various special cases and 
conditions. It cannot, however, be applied to the moving 
load on skew bridges. The reaction for the centre line of 
the track must be reduced to the trusses, while the negative 
moments of the loads in advance of the panel point which 
serves as the moment origin remain unchanged. The gener¬ 
ality of this method is not, therefore, complete. 


The Maximum Floor-beam Reaction . 

The moving load iscarried to each transverse floor beam by 
the adjacent stringers. Hence each floor beam is a pier for 
two adjacent spans of stringers, and it becomes necessary to 
determine that position of the moving load on those two 
spans which will subject the floor beam to its greatest 
load. 

In the Fig. let a section of the beam be shown at R , while 
/ and T are the two adjacent stringer spans traversed by the 
moving load ; then let the x's be measured from the right 
and left ends of / and while W , W 1 , etc., W x% W 2 , etc., rep¬ 
resent the weights or wheel concentrations resting in the two 
spans, the reaction R will then have the value: 

D W x Xi + W 2 x 2 + etc. W x 4- W x x x + etc. 

R = ~ - 7 - + - 1 - 

If the whole system of loading move to the left by the 
distance ax, the new reaction will be: 





36 SPECIAL NON-CONTINUOUS TRUSSES. 

(W + W 9 4- etc.) A* (W + W 1 4- etc.) A;r 
7 ?=^-jr + / 


In that position which gives a maximum or minimum, 
R t — R — o; hence : 

(J^i + IE 2 + W t + etc.) j - ( W + W l + IV" + etc.) (20). 


It will seldom happen that Eq. (20) will be satisfied unless 



r 



etc. 



etc. 


1 



_l 


L 




Fig. 7. 


a concentration rest on the point R, so that the proper por¬ 
tion of it may be taken for one span or the other, precisely 
as in the problems of maximum shear and maximum mo¬ 
ments. 

Ordinarily the two adjacent spans are equal, or 
/= 4 . • . Wi+ W, + etc. = W + W' + etc. . . (21). 


Eq. (21) shows that when the two spans are equal, the 
amounts of load each side of R must also be equal. 

After the proper position of loading has been determined, 
Eq. (19) will give the maximum reaction desired. 

♦ 

Article 8.—Fixed Weight. 

As the weight of a structure forms a very considerable 
portion of its total load, it becomes a matter of importance 













FIXED WEIGHT. 


3 / 


to find at least an approximate value for it, when the moving 
load has been once assumed. 

The weight of ties, guard timbers or rails, rails, spikes, etc., 
may be taken at 350 to 475 pounds per lineal foot of single 
track, and forms an invariable part of the fixed weight, (i. e., 
weight of structure), since it is independent of span, length 
of panel, or depth of truss. With ordinarily heavy traffic 
and standard gauge, 400 pounds per lineal foot is usually 
taken. 

After having fixed the weight of track, the stringers are to 
be designed. A very little experience will enable their 
weight per lineal foot to be assigned in advance, so that the 
total load resting upon them may be used in making compu¬ 
tations. These track stringers are almost invariably plate 
girders, and the main object of the computation is to deter¬ 
mine the area of flange section. If that area, as computed, 
makes the weight of the stringer very different from that 
assumed, it will be necessary to again assume a weight, 
guided by the results of the first computation, and re-calcu¬ 
late for the flange area ; and then repeat the operation until 
sufficient accuracy is obtained. 

The weight of the track stringer thus obtained becomes a 
part of the load sustained by the floor beam. The weight 
per lineal foot of the latter is then to be assumed, and calcu¬ 
lations made and repeated, if necessary, precisely as in the 
case of the stringers. With a very little practice the weights 
of the stringers and floor beams may be so accurately 
assigned in advance, that a re-calculation is seldom or never 
necessary. 

The lateral and transverse systems of bracing should next 
be subject to computation, and as the wind pressure is their 
only load, their own weight does not affect the operations; 
hence no re-calculation will ever be required. 

The remaining calculations are those of the truss proper, 
and the accurate assignment of their own weights presents 
more difficulty than any other part of the operation. Ex¬ 
perience, however, enables even this weight to be quite 
closely taken in advance. For each single track of standard 


33 


SPECIAL NON-CON TIN UO US TRUSSES. 


gauge the weight of two through trusses, designed for the 
moving load taken in Art. 9, together with the lateral sys¬ 
tem, in pounds per lineal foot, may be approximately taken 
at five times the length of the span in feet, or double that 
amount for a two truss double track structure. If this is not 
sufficiently close, one or more re-calculations will be required. 
For spa©s over two hundred and fifty feet in length, this rule 
gives too small results. 

By thus designing the floor system, and lateral and trans¬ 
verse bracing before the computations are made for the 
trusses, it is only necessary to assign before each step, the 
weight of that part immediately under consideration, and, 
hence, enables the actual weight to agree very closely with 
the assumed, without re-calculation. 

If the bridge is of the deck variety and carries the ties di¬ 
rectly pn the upper chord, so that they act as a transverse 
load on the latter, thus obviating a system of track stringers 
and floor beams, the total fixed weight will be reduced about 
125 pounds per lineal foot. If, on the other hand, such a 
bridge carries a regular system of stringers and floor beams, 
the truss weight may be the same as if it were a through 
bridge. 

If the span is less than about eighty or more than about two 
hundred and fifty feet, the weight per lineal foot will some¬ 
what exceed the value given by the preceding rule. 

If the bridge is a highway structure, the same general 
method of operations (and taken in the same order) is to be 
followed as for a railway bridge. A rule for truss weights can¬ 
not, however, be so easily given, because the moving load is 
so very variable; they may equal or exceed those of railway 
structures of the same span, or may not exceed a third of that 
value. The weight of a highway floor, if of plank and timber 
joists, will be from twenty to twenty-five pounds per square 
foot. 

Art. 9.—Single System of Bracing with Two Inclinations. 

The first case taken will be that shown in Fig. 2 of PI. I. 
The span s is 120 feet; depth d 20 feet; panel length p 20 


RAIL WA Y BRIDGES. 


39 


feet; angle NAR , i8° 30', and angle NAM, 33 0 40'. The 
moving load will be taken as two coupled consolidation loco¬ 
motives, each with weights distributed as shown in Fig. 1, 
Art. 77, followed by a uniform load of 1.5 tons per lineal foot. 
The bridge will be supposed to be a single track “ through ” 
structure. Each truss will be taken to weigh approximately 
240 pounds per lineal foot. The upper lateral bracing will be 
taken at thirty pounds per lineal foot for each truss. The 
total weight concentrated in each of the upper chord panel 
points will then be 10 x 270 = 2,700 pounds. The fixed 
weight concentrated in the lower panel points will be taken 
at 9,300 pounds. Hence, if W' is the upper chord fixed panel 
load and W the same for the lower: 

W' — 2700 pounds — 1.35 tons, for one truss. 

W = 9300 “ = 4.65 “ “ “ “ 

tan NAR = 0.333 l an NAM — 0.666 

sec “ = 1*054 sec “ = 1.202 

In the above fixed weight, the ties, rails, guard timbers, etc., 
were taken at 400 pounds per lineal foot. 

The stresses due to the fixed load only, in each of the truss 
members will first be found, and it will be convenient to begin 
by determining those in the web members. 

On page 7 it is shown that the stress in any web member 
of a truss with horizontal chords is the vertical shear, multi¬ 
plied by the secant of its inclination to a vertical line. But 
the vertical shear in any web member of such a truss is simply 
the algebraic sum of all the vertical forces or weights ( includ¬ 
ing the end reaction ) between the end and the web member in 
question. 

In web member 5, for example, the fixed load shear will be 
the difference between the reaction R and the weights at A , 
B, M and L\ or, since the truss is symmetrical with the 
centre, the shear will be the weight at C added to half the 
weight at K. In fact, as a general principle, when the truss 
and its load are symmetrical with the centre, the shear in any 
web member will be the load between that member and the 


40 


SPECIAL NON-CONTINUOUS TRUSSES. 


centre . Hence, the shear in the web member of half the truss 
will possess the following values : 

Shear in 6 - \W = 2.325 tons 

“ “5=1 W + W' — 3.675 “ 

“ “4=1 \W + W' = 8.325 “ I 

“ “ 3 = iifT+2PF'= 9.675 “ [ ‘ * U ‘ 

“ “ 2 = 2 \W + 2 W' — I4.325 “ 

“ “ I = 2\W + 3W' = 15.675 “ 


If the plus sign indicates tension and the minus sign com¬ 
pression, the web stresses will take the following values: 


Stress in 6 = + 2.325 x 1.202 = + 2.79 tons 

“ “ 4 = + 8.325 x “ = + 10.00 “ 

“ “ 2 = + 14.325 x “ = + 17.19 “ 

“ “ s = — 3- 6 75 X 1-054 = - 3-87 “ 

“ “ 3 = — 9.675 X “ = — 10.18 “ 

“ “ I = — 15.675 X “ = — 16.52 “ 



On the same page, 7, it was shown that the increment of 
chord stress at any panel point is equal to the algebraic sum 
of the horizontal components of the web stresses intersecting 
at that point; but those horizo 7 ital components are the vertical 
shears multiplied by the tangents of the respective inclinations 
to a vertical line. Hence: 


The chord increment at: 


A = 15.675 x} + i 4 - 3 2 5 x | = 14.79 tons 
B = 9.675 x 1 + 8.325 x | = 8.79 “ 

C = 3.675 x 1 + 2.325 x | = 2.79 “ 

R = 15.675 x 1 + o = 5.24 “ 

M= 9.675 x l + 14.325 x | = 12.79 “ 

L — 3.675 x 1 + 8.325 x | = 6.79 M 

K — O + 2.325 x | = 1.55 “ 




• (3)- 


J 


Hence the following are the upper chord stresses: 

(1) = — 14.79 = — 14-79 tons ) 

(2) = - (14.79 + 8.79) = - 23.58 “ l . (4). 

(3) = - (23.58 + 2.79) = - 26.37 “ ) 


The lower chord stresses wiil take the values: 







RAILWAY BRIDGES. 


41 


(1) = 5.24 tons \ 

(2) = 5.24 + 12.79 = 18.03 “ . (5). 

When the position of the moving load is once determined for 
a required maximum stress, the latter is found from the re-ac¬ 
tion and panel loads by precisely the same general methods 
used with the fixed loads. Hence the proper position of the 
moving load is first to be found for each of the web stresses. 

Eq. (7) of Art. 7 is an expression of the condition which 
obtains with the greatest shear in any web member. The 
number of panels, n , is 6. W lt W 2} W 3 , etc., are the single 
locomotive weights given in Art. 77; W\ has the value 3.75 
tons; W 2 , W 3 , and W 5 , 6 tons, etc., etc. 

If the moving load passes on the bridge from the right, and 
rests at the foot of web member 10, W t . . .. W 5 will 

be on the bridge, and n times 375 ( i.e .,6 x 3.75 = 22.50 tons) 
v/ill be less than W t + W 2 + . . . + W s = 27.75 tons. But 

6 x (3*75 + 6) tons is much greater than the total moving 
load on the bridge. Hence, W 2 at G is the position desired 
for web member 10 ; will then be 8.083 feet from G toward 
H. At this point the tabulation mentioned in connection 
with Eqs. (8) and (15) of Art. 7, may be used. 

The tabulation for the two locomotives is given herewith. 


I 




3 

4 

I 

15,000 

X 

8.08 

121,200 

121,200 

2 

39,000 

X 

5-75 

224,250 

345,450 

3 

63,000 

X 

4 50 

283,500 

628,950 

4 

87,000 

X 

4-50 

391.500 

1,020,450 

5 

111,000 

X 

7.08 

785,880 

1,806,330 

6 

126,000 

X 

4-83 

608,580 

2,414,910 

7 

141,000 

X 

5-67 

799 > 47 ° 

3,214,380 

8 

156,000 

X 

4-83 

753 . 48 o 

3,967,860 

9 

171,000 

X 

9.00 

1,539,000 

5,506,860 

10 

186,000 

X 

8.08 

1.502,880 

7,009,740 

11 

210,000 

X 

5-75 

1,207,500 

8,217,240 

12 

234.000 

X 

4 50 

1,053,000 

9,270,240 

13 

258,000 

X 

4-50 

1,161,000 

10,431,240 

14 

282.000 

X 

7.08 

1,996,560 

12,427,800 

15 

297,000 

X 

4-83 

1,434,510 

13,862,310 

16 

312,000 

X 

5-67 

1,769,040 

15,631.350 

1 7 

327,000 

X 

4-83 

i, 579 . 4 io 

17,210,760 

18 

342,000 

X 

4.00 

1.368,000 

18,578,760 










42 


SPECIAL NON-CONTINUOUS TRUSSES. 


A little consideration of the table will make its composition 
evident. It is reproduced from a table in actual use, and is 
given in pounds and foot pounds. 

When W 2 rests at G, the x of Eq. (8) will be 5- 2 5 feet. 

It should be explained that each quantity in column 4 is 
the sum of all the preceding and opposite numbers in column 
3. As an example, 5,506,860 is the sum of all the numbers 
in column 3 from the top down to and including the ninth. 
Column 4, then, represents the positive parenthesis in Eq. 
(8) of Art. 7, less the term multiplied by ;r. 

When W 2 rests at G , the x of Eq. (8), Art. 7, will be 5.25 
feet. Hence that equation in connection with the preceding 
table gives : 


. 1 1 11000 

Shear in 10 = -(- 

120 \ 2 


x 5.25 + 


1020450^ 


1 121200 

— x-. 

20 2 


= 3650 pounds — 1.825 tons.(6). 

This quantity will shortly be needed again. 

If the same locomotive weight W 2 rest at H, W x . . . 

W g , or one complete locomotive, will be found on the bridge, 
and n times (3.75 + 6) or 6 x 9.75 = 58.50 tons is still in excess 
of 42.75 tons, i. e., half the entire locomotive weight. Hence, 
W 2 at H is the position of moving load, which gives the 
greatest shear to web member 8, and x, in Eq. (8) of Art. 7, 
becomes 2.83 feet. That equation then gives : 

Shear inZ = A- x 2.83 + - I x ^°°. 

120 \ 2 2 / 20 2 

= 15520 pounds — 7.76 tons.(7). 

If the same weight, W 9 , be placed at K , it will be found that 
W 12 will rest at the right extremity of the span,, there will 
therefore be eleven weights on the bridge. Since 6 x (12000 
+ 7500) = 117000 > 105000, this position of the moving load 
gives the greatest shear in brace 6. Eq. (8) of Art. 7 then 
gives, since x — 5.75 feet: 

^ 8217240 1 121200 

Shear in 6 = -— -x - 

2 x 120 20 2 

= 31400 pounds = 15.7 tons 


( 8 ). 













RAIL IVA V BRIDGES. 


43 


If W 2 be placed at Z, it will be found that six times 
(W 1 + Wj) is less than W 1 . . . . W 15 , which will then 
rest on the bridge; but 6 ( W 1 + W 2 V JV 3 ) > {W 1 . . . . 
JV 1G ), hence W 3 must rest at Z for the greatest shear in 4. 
The value of .r will then be 4.83 feet. Hence : 

Shear in 4 = — (^°- x 483 + - ± x M 1450 

120 \ 2 2 / 20 2 


— 5548 0 pounds — 27.74 tons.(9). 

In the case of brace 2, the condition of maximum shear 
obtains with W 3 at M. This position of moving load places 
10.5 feet of the uniform load of 1.5 tons per lineal foot on 
the bridge. The value of x in Eq. (8) of Art. 7, conse¬ 
quently, becomes 5.25 feet. The tabulation given above 
is not quite sufficient to completely cover this case; although 
the result in line 18 forms by far the greater portion of the 
moment product which must be divided by the span, in 
order to obtain the shear. The application of Eq. (8) of Art. 
7, to this particular case becomes, then: 


r , . 1 /342000 , 3000 „ . t 

Shear in 2 ==- (—- x 10.5 + x 10.5 x 5.25 f 

120 \ 2 J 2 

1857876CA _ 345450 _ 

2 / 2 X 20 * 

= 84430 pounds = 42.2 tons.( I0 )- 

The moving load shears in web members 1, 3> 5 > 7 an< ^ 9 ’ 
will be precisely the same as those in 2, 4, 6, 8 and 10 re¬ 
spectively, because each pair, as 10 and 9, 8 and 7, etc., inter¬ 
sect in a chord which carries no moving load. The web 
stresses due to the moving load will then take the following 
values: 


Stress 

in 

IO = 

+ 

1.825 

X 

1.054 = 

4 - 

1.92 

tons. 

u 

n 

8 = 

+ 

7.760 

X 

u _ 

+ 

8.1 

<< 

u 

u 

6 = 

+ 

157 

X 

1.202 = 

+ 

18.87 

<( 

u 

u 

4 = 

+ 

27.74 

X 

(( _ 

_U 

1 

33-35 

u 

:i 

«« 

2 = 

+ 

42.20 

X 

u _ 

+ 

50.72 

u 


(11). 


♦ 














44 


SPECIAL NON-CONTINUOUS TRUSSES . 


Stress in 

(< (. 

U il 

il 11 

a a 


9 = — I.825 X 1.202 = — 2.19 tons. ' 

7= - 7.762 x “ = — 9*33 “ 

5 = — 15.7 x 1.054 = — 16.49 “ 

3 =-27.74 x “ =-29.13 “ 

1 = —42.20 x “ =—44.30 “ 



By comparing (n) and (2), it is seen that the moving load 
stress in web member (10) is of a kind opposite to that caused 
by the fixed load in web member 3, while those two members 
are, in reality, identical ; the latter is compression and the 
former tension. Since it is evident that no piece of ma¬ 
terial can be compressed and extended at the same time, it 
is clear that the resultant stress will be the numerical differ¬ 
ence or algebraic sum of the induced stresses. In other 
words : 

If the action of forces external to a piece of material tend to 
subject that portion of material to stresses of opposite kinds , the 
resultant stress zvill be equal to the numerical difference of the 
opposite stresses , and will be of the same kind as the greater. 

By comparing (11) and (12) with (2), it will be seen that in 
all the web members of that half of the truss first traversed by 
the train, the fixed and moving load produce stresses of opposite 
kinds. The moving load stresses predominate in the mem¬ 
bers near the centre of the span, but near the ends the fixed 
load stresses are the greatest. Li web member 8, the resul¬ 
tant stress is 8.1 — 3.87 = 4.23 tons of tension ; but if the 
bridge carries no moving load, it is subjected to 3.87 tons of 
compression. Now, that member may be so designed and 
constructed that it can resist tension or compression, accord¬ 
ing to the demands upon it; in such a case it is said to be conn - 
terbraced. If, however, it is formed to resist compression only, 
the member 14 must be introduced between .//and C to take 
the moving load shear in tension. In order to provide for 
the movement of the load in the opposite direction, 13 must 
be introduced between L and D. Such web members as 13 
and 14 are called coimterbraces. Other web members than 
counterbraces are called main braces or main zveb members. 
The duty of counterbraces , then, or of counterbraced main 



RAIL WA Y BRIDGES. 


45 


web members, is to transfer moving load shear to the farther 
abutment or pier. 

It is now clear that the extent to which a main web mem¬ 
ber must be counterbraced is found by taking the excess of 
the moving load stress over that caused by the fixed load. 
At first sight it would appear that the same method should 
hold in determining counterbrace stresses ; since it may be 
supposed that the main brace will carry shear until its fixed 
load stress is neutralized. This presupposes, however, that 
there is such an exact adjustment of members that each will 
perform just the amount of duty assigned to it. As that is 
an end that can never be confidently realized, it is only pru¬ 
dent to suppose that the counterbrace takes all the moving load 
shear , and this will be assumed in all that follows. This 
procedure appears the more advisable when one reflects 
that counterbraces are subject to the greatest fatigue of all 
truss members, and that the amount of metal concerned is 
trifling. 

It now becomes necessary to determine where the coun¬ 
terbraces are to begin. Since the fixed and moving load 
stresses, or shears, neutralize each other in equal amounts, it 
at once results that counterbraces or counterbraced zveb mem¬ 
bers must begin at that point , or with that web member , in 
which the stress or shear produced by the moving load is 
greater than that of the opposite kindproduced by the fixed load 

“ Stress ” or “ shear ” is used indifferently, as the former is 
simply the product of the latter by the secant of the inclina¬ 
tion to a vertical. 

In order, therefore, to find the stress in a counterbrace, it 
is only necessary to ascertain the moving load shear, and 
multiply it by the secant of the inclination. The secant of 
the angle between counterbrace 13 or 14, and a vertical line 
is 1.944; and since the shear it has to carry is by Eq. (11) 7.76 
tons: 

Stress in counterbrace (14)=7.762 x 1.944=T1 5-°9 tons. (13). 

Again, by comparing (n) with (2) it is seen that the fixed 
load shear in 10 (or 3) is — 9.675 tons, while the moving load 


4 6 


SPECIAL NON-CONTINUOUS TRUSSES. 


shear is + 1.825 tons. Hence 13 and 14 are the only counter¬ 
braces needed. 

It is farther seen that in the half of the truss traversed last 
in order by the train, the stresses produced by the fixed and 
moving loads are the same in kind. Hence, in the main 
braces the fixed and moving loads induce stresses of the same 
kind and the resultant is simply the numerical sum . 

A tabulation of all the resultant web stresses, then, gives 
the following values: 


Web member 1 

= — 60.82 tons. 

a 

u 

3 

— — 39 * 3 1 “ 

<< 

u 

5 

— — 20.36 “ 

u 

i( 

2 

= + 67.91 “ 

a 

u 

4 

— + 43*35 “ 

u 

il 

6 

= -f 21.66 “ 

u 

u 

14 

= + 15.09 “ 



The moving load chord stresses remain to be found, and it 
will be necessary to resort to the methods of Art. 7 in order 
to determine the proper positions for the stresses in the vari¬ 
ous panels. 

Since none of the web members are vertical, the positions 
of moving load for the greatest stresses in the lower chord 
panels will be given by Eq. (18) of Art. 7. For this case, q 
in that equation will be one-third the panel length. 

In finding the stresses in lower chord panels 2 and 3, it is 
necessary to bring 10.4 and 7 feet, respectively, of the uni¬ 
form load on the bridge. The condition of greatest stress in 
the lower chord end panel coincides with that for brace 1, 
and it will only be necessary to multiply the greatest shear 
in that brace (already determined) by the tangent of its in¬ 
clination to a vertical line. 

In determining the greatest stresses in the upper chord, 
moments are taken about the lower chord points, and Eq. 
(14) of Art. 7 will be used. 




RAIL IV A Y BRIDGES. 


47 


The application of Eqs. (18) and (14) of Art. 7 give the 
following results : 


/' 


Lower cli a 

7 2 . 

' / 

— _ 4 _ 

— T 8 • • 

Wi 

= w t . 

.x x — 6.67 ft . 

.2 


IO.4 

u 

u 

3 • 

U 

• 

= JL 
— 1 8 * • 

u 

= W%. 

M 

A 

II 

• 

.2 


7.0 

Upper 

u 

I . 

u 

• 

— 1 
— J ■ • 

w n „ 

-1 = W, . 



X — 

IO.4 

a 

u 

2 . 

u 

• 

- 1 

— 3 * • 

u 

= w, . 


. 2 

X = 

6.4 

u 

u 

3 • 

u 

— 1 

2 * * 

u 

= W U . 


..2 

X — 

I9.O 


a 

u 


Eq. (16) of Art. 7, applied to the lower chord, gives by the 
aid of the tabulation already employed the moments: 


In lower chord 2 ; M — A 
x 1500) 5.2 


18,578,760 


+ (2 x 171000 + 10.4 


_ 34545 9 _ 63 000 x fi fty _ £ ( I2 000 X 15.5 
+ 12000x11 + 7500x3.92), 


In lower chord 3 ; M — T V 


.*. M— 1,936,400 ft. lbs. 

18,578,760 


x 1500)3.5 


1,806,330 126000 


+ (2 x 171000 + 7 


x 7.2 — J (7500 X 15.67 


+ 2 X 7500 X 7.58), 

.*. M— 2,658,490 ft. lbs. 

Eq. (15) gives for the upper chord moments: 

18,578,760 


In upper chord 1 ; M — \ 


x 1500) 5.2 


+ (2 x 171000 + 10.4 


34545Q , 
2 


•\ M— 1,685,425 ft. lbs. 



















48 


SPECIAL NON-CONTINUOUS TRUSSES. 


In upper chord 2 ; M — \ 


18,578,760 


+ (2 x 171000 + 6.4 


x 1500) 3.2 


1,806,330 


.*. M — 2,568,335 ft. lbs. 


In upper chord 3 ; M — \ 
x 1500) 9.5—7500x 122.5 


18,578,760 


+ (2 x 171000 + 19 


'8,217,240 


— 7500 x 62.5 


M= 2,305,315 ft. lbs. 

The negative moments ( — 7500 x 122.5) an d (—7500 x 62.5) 
occur in the last moment above, for the reason that W x is 
found 2.5 feet off the span to the left when upper chord 3 
takes the maximum bending. 

It is notable that the moment in upper chord panel 3 is 
less than that in panel 2, while it is yet more important to 
observe that the uniform load of 1.5 tons per lineal foot gives 
the greatest bending moment in the centre panel of the upper 
chord. A panel uniform load is 10x1.5 = 15 tons = 30000 
pounds, and the reaction with this load on the whole bridge 
is 2.5 x 15 = 37.5 tons=750oo pounds. Hence, making K the 
origin of moments: 

hi upper chord 3 ; M = 75,000 x 60 — 2 x 30,000 x 30=2,700,000 

ft. lbs. 

These operations verify a previous observation to the 
effect that with concentrated loads there may be several 
maxima. 

Eq. (10) shows that the greatest moving load shear in 
braces 1 and 2 is 42.2 tons; hence, multiplying that result by 
its tangent, 0.333, and dividing the preceding greatest mo¬ 
ments by the depth of truss, i.e.> 20 feet, the following mov¬ 
ing load chord stresses are found: 










RAIL IVA Y BRIDGES. 


49 


In upper ch'd i ; - 5,425 = 84,271 lbs. = 42.14 tons. 


<( (( u 


u u u 


2; - 


3; 


20 

* 

2,568,335 

20 

2,700,000 

20 


= 128,417 “ =64.21 “ 


= 135,000 “ =67.50 “ 


In lower did 1;— 42.2 x 0.333 =14.06 “ 

1,936,400 


u u u 


2; - 


20 


= 96,820 lbs.= 48.41 “ 


u a 


2,658,490 „ 

3 ;-^-^-= 132,924 “ =66.46 “ 


05 )- 


The resultant chord stresses are found by adding groups 
(4) and (5) to (15), as follows : 


Upper chord (1) = -(14.79+42.14) = — 56.93 tons. 

“ “ (2) = -(23.58 + 64.21) =—87.79 “ 

“ “ (3) = -(26.37+67.50) = -93.87 “ 

Lower chord{\) = +( 5.244-14-06) = + 19.30 tons. 
“ “ (2) = +(18.03+48.41) = +66.44 “ 

“ “ (3) = +(24.72+66.46) = +91.18 “ 



Groups (14) and (16), therefore, give the resultant maximum 
stresses in all members of the truss. 

Those web members, such as I, 3 and 5, which sustain com¬ 
pression, are called “posts ” or “ struts,” while those, such as 
2, 4, 6 and 14, which sustain tension, are called “ ties.” 

If the truss be divided through CD and either LK or KH, 
it is seen that more than three members must be cut; but if 
that number is exceeded, it is known from the first principles 
of statics that the stresses must become indeterminate. 
Hence, when counterbraces are introduced , indetermination 
always results. If provision is made for one system of legiti- 









50 


SPECIAL NON-CON TIN UO US TRUSSES. 


mate stress analysis, however, the safety of the structure is 
assured. 

Only one point more needs passing attention before the ex¬ 
amination of the next case. It has been stated in the course 
of the demonstrations that the stress in certain members is 
tension, and compression in others.♦ In web member 4, for 
example, let it be desired to determine the kind of stress. It 
has been seen that when the greatest main web stress exists 
in that member, the reaction at R is 32.058 tons, and it is 
evident that it is directed upward . At the same time the 
live load resting at M is 4.32 tons and is directed down . The 
difference of these forces is an upward shear of 27.74 tons. 
H ence, if the truss is divided anywhere between BM and CL , 
this shear will tend to move the left portion (between the 
line of divisions and R) upward and past the right portion; 
i. e. y it will tend to increase the distance between B and Z, 
and, consequently, produce tension in web member 4. The 
general principle then is to determine the effect of the re¬ 
sultant external forces on the distance between the extrem¬ 
ities, or any other two points in the axis of the member; if 
the tendency is to increase this distance, the resulting stress 
will be tension, and compression if the reverse is the case. In 
trusses with parallel chords, after a very little experience, the 
kind of stress in any member may readily be discovered at a 
glance, but ‘in many structures with curved or polygonal out¬ 
lines, resort must be made to the general principle stated 
above, which will be more thoroughly given hereafter. This 
simple statement, however, is all that is needed here. 

Art. 10.—Single System of Vertical and Diagonal Eracing.—Verticals 

in Tension. 

This form of truss when built with timber compression 
members, has long been known as the Howe truss. The skel¬ 
eton diagram of the structure to be considered is shown in 
Plate I, Fig. 3. The moving load will be supposed to pass 
along the upper chord; hence the bridge is a “ deck” struc¬ 
ture. The following are the principal dimensions and fixed 
load data: 


RAIL WA Y BRIDGES. 


51 


Span = 98 feet. Panel length = 14 feet. 

Depth = 20 feet. Number of panels = 7. 

Upper chord fixed load = 385 lbs. per lineal foot per truss. 

Lower “ “ “ = 215 “ “ “ “ “ “ 

Upper “ “ “ per truss panel = W' = 2.70 tons. 

Lower “ “ “ “ “ “ = W = 1.50 “ 

tan ABC — 0.7 sec ABC — 1.22. 


The moving load will consist of the two consolidation loco¬ 
motives used in the preceding Art., the weights of which are 
shown in Art. 77 ; and this load will be taken as passing 
from N towards M. 

The web stresses will first be determined, and the first 
counterbrace needed comes first in order. As the number of 
panels is seven, n in Eq. (7) of Art. 7, is equal to 7. Let it 
be required to ascertain whether the compression counterbrace 
LK is necessary. If the train is so placed that W 2 rests at Z, 
the bridge will carry the weights ( W l ••• W 7 ) = 70.5 tons, or, 
35.25 tons on each truss. Now 7 W 1 = 52.5 tons, and 7 (W l 
_j_ W 2 ) — 136.5 tons. Since the total load on the bridge is 
found to lie between these values, by the principles of Art. 7 
it is placed to give the greatest compressive shear in KL or 
tensile shear in HP. In order to find the shear by Eq. (8) of 
Art. 7, the following values result from the position of the 
moving load just taken; = 1.3 ft., W n = W 7 and W n >_ t = 
W l . Since l = 98 and p — 14, Eq. (8) of Art. 7 gives by the 
aid of the tabulation on page 41 ; 




> 4 I 4 > 9 I ° 

2 


+ 


141,000 

- x 

2 



1 121,200 

— X - 

14 2 


= 8920 lbs. — 4.46 tons. 


The fixed load shear in the same panel is W' + W = 4.2 
tons. As the latter is less than, and opposite in kind to that 
of the moving load, the counter post or strut KL must be 
introduced. Since the difference in these shears is very small, 
it is evident that no counterbrace between KL and O is needed, 
and that conclusion may easily be verified. 





52 


SPECIAL NON-CONTINUOUS TRUSSES. 


By proceeding in precisely the same manner for the shears 
in the other panels, the following quantities are found for in¬ 
sertion in Eq. (8) of Art. 7, when Eq. (7) of that Article is 
satisfied; 

/ 

For greatest shear in 

KL . . . W 2 at L . . . W n = W 7 . . . x — 1.3 . . . W n ._ x - W v 

HG... W 2 “ H... W n = W, . . . x = 4.8 . . . W n ._ x = W t . 

FE.. . W 2 “ F .. . W n = W n . . . x = 1.8 . . . W n ._ x = W x . 

DC .. . W 3 “ D ... W n = W u . . . x = 6.8 ... W n ._ x = W* 

BA ... W 3 “ B . . . W n = W 17 . . . x = 2.5 . . . = J>F 2 . 


That Eq. (8) then gives : 

- 2,414,910, 141,000 

Shear in KL = —_ 

98 


x 1.3 


1 121,200 

— x -— 4.46 tons. 

14 2 


_ 1 r 3,967,860 171,000 n _ 

“ “HG= I - v -+——'X4.8 

98 j 2 2 

« “ FE= - 7 ’ 009 ’ 74 ° + —’°^ 0 x i.8* 

2 2 


1 121,200 

— x - 

14 2 


= 10.05 


K 


98 


“ “ DC= - 

98 

** “BA =-\ 
98 


'10,431,240 282,000 

■- — + •— -x 6.8 


15 631,350 ( 327,000 ^ / 

r * X Z, 5 


1 121,200 , 

— x --- = 16.68 

14 2 


1 345450 

- x ----25.33 “ 

14 2 


I x 345 l 45°_ .. 

14 2 


The shears in the inclined web members only have been 
given, because each pair of braces that intersect in the chord 
not traversed by the moving load take their greatest stresses 
with the same position of moving load. Braces 2 and 3, 4 
and 5, etc., thus go together in pairs. 

The resultant web stresses, by the aid of the preceding 
results, will then take the following values: 

Brace 1 . . . —[3 (W + W 1 ) + 35.79] x 1.22 

“ 3 . . . —[2 (W + W 1 ) + 25.33] x T - 22 

“ 5 . . . — [ W + W 1 + 16.68] x 1.22 

“ 7 . . . — 10.05 x I - 22 

“ 8 . . . — 4.46 x 1.22 


= -- 59.04 tons. 
= - 4 i.i 5 “ 

= - 2 548 “ 

= — 12.26 “ 



























RAIL IVA Y BRIDGES. 


53 


Brace 2 . . . 2 W 1 + 3 W + 25.33 = + 35.23 tons. 

“ 4 . . . W 1 + 2W + 16.68 — + 22.38 “ 

“ 6 . . . W + 10.05 — + n.55 “ 


The positions of the moving load for the greatest chord 
stresses are found by the aid of Eq. (14) of'Art. 7, and result 
in the following quantities : 


For upper chord 

3 . . . Wj at F . . . W n — W 16 ... x — 2 it. 
2 .. . W 6 “ D . . . W n = W 16 . . . x = 4 “ 

1 . . . W, “ B 


W n ,_1 - fH 6 . 

^-1 = ^4. 


The greatest stress in upper chord I occurs with the maxi¬ 
mum shear in BA, and is found by taking the product of that 
shear by the tangent of its inclination to a vertical line. The 
shear has already been determined to be 35.79 tons, and the 
tangent is 0.7. Hence ; 

Stress in upper chord 1 = — 35-79 x 0.7 = — 25.05 tons. 


Eq. (15) of Art. 7 then gives by the aid of the tabulation on 
page 41, the following bending moments: 

13,862,310 312,000 


Upper chord 3 =r_ 
7 

2 

“ “ 2 —— 

7 


x 2 


'13,862,310 312.000 

—-——+ - x 4 


2,414,910 


= 1,896, 754 ft. lbs. 


1,020,450 


= 1,648,391 


tC U 


As the depth of the truss is 20 feet, the moving load chord 
stresses become: 

Upper (3) = — 1,896,754 ~ 20 = - 94,837 lbs - = — 47 A 2 tons. 
“ (2) = — 1,648,391 -f- 20 =— 82,420 “ =—41.21 “ 

“ (1)= =-25.05 M 

By combining these results with those due to the fixed 
load, the following resultant chord stresses are found : 

Upper (1) = - [25.05 + 3(^+ W 1 ) x 0.7] = - 33.87 tons. 

“ (2) = - [41.2 1 + 5 (W 4- W') x 0.7] = - 55.91 “ 

“ (3) = - [47.42 + 6( W + W l ) x 0.7] = - 65.06 “ 












54 


SPECIAL NON-CONTINUOUS TRUSSES. 


It is to be observed that the upper and lower chord stresses 
are the same in pairs, i. e., in the same oblique panel. The 
reason is obvious. If a panel, as DFEC , be divided by any 
line cutting DF, DE and EC, it will be evident that no hor¬ 
izontal forces whatever exist except the stresses in DF and 
EC\ hence, by the first principles of statics, those stresses 
must be equal in amount and opposite in kind. The same 
result holds for any oblique panel, and, indeed, in all vertical 
and diagonal bracing with horizontal chords and vertical load¬ 
ing. The stresses in one chord can then always be written 
from those in the other, as is done here: 

Lower (i) = + 33.87 tons. 

“ ( 2 ) = + 55-9 1 “ 

“ (3) = + 65.06 “ 

In reality the moments remain the same whether the upper 
or the lower extremity of any vertical brace is taken for the 
moment origin ; hence the equality of upper aud lower chord 
stresses in the same oblique panel. 

If the truss becomes a through one (i. e., with the load on 
the lov:er chord), the same pairs of web members do not inter¬ 
sect in the unloaded chord, hence a different position of the 
moving load must be taken for the greatest shears in half the 
braces. A simple inspection of the diagram will show at 
once that the position of the moving load for the greatest 
shears in the oblique or compression braces must be the same 
whether that load traverses the upper or lower chord. For 
the vertical braces, however, the moving load must be ad¬ 
vanced in the lower chord at least one panel (more in some 
cases) beyond its position on the upper. Hence the vertical- 
brace stresses will be greater in a through truss than those 
found with the moving load on the upper chord, while the 
stresses in the oblique braces remain the same. Conse¬ 
quently this type of truss is better adapted to the deck than 
the through form. 

Art. 11.—Single System of Vertical and Diagonal Bracing.—Verticals in 

Compression. 

This type of truss is very common in American bridge 


RAIL WA V BRIDGES. 


55 


practice. Its compression members are the shortest possible, 
and its details are simple in character, and both those fea¬ 
tures are conducive to economy. The skeleton diagram to 
which reference is to be made is given by Fig. i, of Plate II. 
The bridge is supposed to be a “through” structure, hence 
the floor system will rest on the lower chord. The loads and 
stresses in this Art. will be given in pounds. The principal 
dimensions and fixed load data are as follows: 

Span = 184' iif” Panel length = 20' 6f” 

Depth = 27' o” Number of panels = 9 

Upper chord fixed load = 230 lbs. per lin. ft. per truss. 
Lower “ “ = 533 “ “ “ 

Upper “ “ per truss panel = W' — 4726 lbs. 

Lower “ “ “ “ =W = 10953 “ 

15679 lbs. 

tail ABL = 0.761 sec ABL — 1.26 

The moving load will consist of a train of two coupled con¬ 
solidation locomotives, each with concentrated weights as 
shown in the diagram below, followed by a uniform train of 
2,240 pounds per lineal foot. It will be taken to move from 
right to left. 

The moving load is comparatively light, hence the fixed 
load is also assumed rather low. 

As usual, Eq. (7) of Art. 7 will be used in determining 



the positions of moving loads for the greatest shears, and, 
hence, for the greatest stresses. For a reason that will here¬ 
after appear, the greatest shear for every panel in the truss 
will be found. The application of Eq. (7), Art. 7, will give 











56 


SPECIAL NON-CONTINUOUS TRUSSES. 


the following quantities for use in Eq. (8) of the same 
Art.: 


Max. shear 

in 





OP . . 

. W x at P .. . W n = W t 

• • • 

-T = 

4.1 

ft. 

MN. . 

. W, “ N . . . “ = W, 

• • • 

X — 

2.2 

U 

FG . . 

G “ = W i% 

• • • 

X — 

1.0 

u 

EH. . 

. “ “ H. . . “ = W a 

• • • 

X = 

2.0 

u 

FI . . 

.W % “ I .. . “ =W) 


2 X — 

9.O 

a 

CJ . . 

. “ “ J . . . “ =w 

Uniform 

2 X — 

29.5 

u 

• BK . . 

. “ “ K . . . “ = W 

load. 

2 X = 

50.2 

u 

AB . . 

. “ “ L . . . “ =W 

J 


2 X — 

70.75 

ii 


The tabulation to be used in connection with Eq. (8) and 
(15) of Art. 7, is given below and is formed precisely like 
that in Art. 9. 


I 

12,000 

X 

7.666 

92.000 


2 

32,000 

X 

4-583 

146,666 

238,666 

3 

52,000 

X 

4-25 

221,000 

459,666 

4 

72,000 

X 

4-583 

329,976 

789,642 

5 

92,000 

X 

10.666 

9 8l >332 

1,770,974 

6 

106,000 

X 

4.666 

459,333 

2,230,307 

7 

120,000 

X 

5.583 

670,000 

2,900,307 

8 

134,000 

X 

4.666 

625,333 

3,525,640 

9 

148,000 

X 

9-333 

i, 38 i ,333 

4,906,973 

10 

160,000 

X 

7.666 

1,226,666 

6,133,639 

11 

M 

OO 

O 

V* 

O 

O 

O 

X 

4-583 

825,000 

6,958,639 

12 

200,000 

X 

4-25 

850,000 

7,808,639 

13 

220,000 

X 

4-583 

1,008,333 

8,816,972 

14 

240,000 

X 

10.666 

2,560,000 

11,376,972 

15 

254,000 

X 

4.666 

i,i 85,333 

12,562,305 

16 

268,000 

X 

5-583 

1 , 496,333 

14,058,638 

17 

282,000 

X 

4.666 

1,316,000 

15 , 374,638 

18 

296,000 

X 

3-25 

962,000 

16,336,638 


Eq. (8) of Art. 7 gives by the introduction of the quanti. 
ties found above, and by the aid of the tabulation: 


Shear in OP = 


4 * “ MN = 


185 

1 


459,666 72,000 


EG 


185 

1 


2 2 

3,525,640 148,000 


x 4.1 


x 2.2 


= 2,040 lbs . 


92.000 
2 x 20.55 


= 8,171 


185 


6,958,639 200,000 

J - +---X 1.0 


= 17,110 


/ 


























RAIL WA Y BRIDGES. 


57 


Shea7 in EH — 

“ “ DI — 

** “ cy 

“ “ BK 


185 

1 

^85 

1 


Ho76,972 , 254,000 

-b -- X 2.0 

2 2 


92000 


16,336,638 6l2,l6o 


X 4-5 


2 X 2O.55 

238,666 

2X 20.55' 


185 

I 


I85 


,, 658,080 

+ - 2 -T-X14.75 

( ( , 704,448 




1 

185 


2 

750,480 


x 25.1 
X35-38 


29,884 
45,788 “ 

= 63,735 “ 

= 86,132 “ 


a 


• IIO, 105 


<6 


Since the fixed load shear in HM is 15,679 lbs. it is seen 
that counterbrace 10 is the first counter needed. Combining 
the fixed load shears with those above due to the moving 
load, the following resultant stresses in the inclined web mem¬ 
bers are found : 

Stress (10) = +(17,110 ) x 1.26=:+ 21,560 lbs. 

“ (9) = + ( 29,884 ) x “ = + 37»654 “ 

“ ( 7 ) = + ( 45>788 + W + W) x “ =+ 77,445 “ 

“ (5) = 4 - ( 63,735 + 2 W+ 2 W') X “ =+119,817 “ 

“ (3) — + ( 86,132 + 3 W + 3 W) x “ = + 167,800 “ 

“ (1) = — (110,105 + 4IV + 4IV') x “ =— 217,750 “ 


Eq. (21) of Art. 7 shows that if be placed at the feet of 
the vertical brace 2, that member will take its maximum 
moving load stress, which, by Eq. (19) of the same Art. takes 
the value: 


(2) = + 


4.05 x 6,000 + (11.72 + 16.32 + 15.97) 10,000 + 2.92 x 14,000 

20.55 

+ 10,000 = + 34,600 lbs . 


The stresses in the vertical braces become: 

Stress (8) = — (29,884 + W) — — 34,611 lbs. 

“ (6) = — (45,788 4- W + 2 W) = — 66,193 “ 

“ (4) = - (63,735 + 2W + iW) =-99.819 “ 

“ (2) = +(34,600 + + W) = + 45.553 “ 

These complete the greatest stresses in the braces. 

























58 


SPECIAL NON-CONTINUOUS TRUSSES. 


Eq. (14) of Art. 7 gives the following values for the posi¬ 
tions of moving load for the greatest stresses in the chords: 

Upper 3 and\ . . . W n < _ t = W u . . . W n =? 2,800 . . . 2 x =65. ft. 

“ 2... “ = w 9 ... “ =82,300... “ =73.5 “ 

“ 1 . . . “ = W S ... u =78,000 . .. “ =69.7 “ 


The “ 2x ” shows the distance covered by the uniform load. 
Eq. (15) of Art. 7 then gives the following maximum bend¬ 
ing moments by the aid of the tabulation already given: 


For upper 3 and 4 


M=- 


16,336,638,737,600.■ 

- i - x 32.3 


u u 


^, 958,639 _ 5)47 g,153//. lbs. 


M = ± 


16,336,638 + 756,640 ^ g - 


4,906,973 


= 4,910,013 ft. lbs. 


u <e 


M 


46,336,638 , 748,130 / 

-o x 34-9 


1,770,974 


= 3.830,813//. lbs. 


The moving load stresses in lower chord 1 and 2 are found 
by taking the product of the greatest shear in brace 1 by its 
vertical tangent: 

Lower (1) and (2) = 110,105 x 0.761 = 83,790 lbs. 


Since the depth is 27 ft. the greatest moving load upper 
chord stresses are the following: 

Upper (3) and (4) =- 5,478,153 -f- 27 = - 202,900 lbs. 

“ (2) =-4,910,013 -f- 27 =- 181,850 “ 

“ (0 =~ 3.830,813 ~ 27 =- 141,900 “ 

Since (W + W') tan ABL— 11,932, the resultant upper 

chord stresses become: 

















RAIL WA Y BRIDGES. 


59 


Upper (i) =-(141,900 + 7 x 11,932),= -225,424/fo. 
( 2 )(181,850 4- 9 x 11,932) =—289,238 “ 

“ (3) and (4) = — (202,900 + 10 x 11,932) =— 322,220 “ 

In the lower chord, the resultant stresses are : 

Lower (1) and (2) = + 83,790 + 4 x 11,932 =+ 131,520 lbs. 

( 3 ) = — 4 225,424 “ 

( 4 ) = = 4 - 289,238 “ 

( 5 ) = =+322,220 “ 

These values complete the resultant stresses in the various 
members of the trusses, and they will be used hereafter in 
making the complete design of this bridge. 

If the moving load traverses the upper chords of the 
trusses, the same pairs of braces as before will not take their 
greatest shears together. The stresses in their inclined braces 
will not in any way be changed, but it will be necessary to 
advance the moving load by at least one panel beyond the 
positions taken in the through bridge, in order to determine 
the greatest shears in the vertical braces of the deck truss. 
Hence, by changing the bridge from the “through” type to 
the “ deck,” the vertical braces will carry considerably in¬ 
creased stresses, and as they are in compression the weight 
of the bridge will be materially increased. Hence, this truss 
is best adapted to carrying the moving load along its lower 
chord. 

Art. 12.—Two Systems of Vertical and Diagonal Bracing.—Verticals in 

Compression. 

The style of truss shown in Fig. 2, PI. II., next to be 
treated, was at one time more common than any other in 
American bridge practice. With increased facilities for fab¬ 
ricating and handling large bridge members, it has been pos¬ 
sible to extend the use of single systems of triangulation to 
much longer spans than formerly. In this manner the am¬ 
biguity of the double system is avoided, and thus analytical 
excellence is combined with advantages of production. In 


6o 


SPECIAL NON-CONTINUOUS TRUSSES . 


long spans, however, the double system is still frequently 
used, and if properly designed it is not so objectionable as 
might at first seem to appear. 

It should always be arranged with an even number of 
panels, as the stress ambiguity is then reduced to an unim¬ 
portant matter. In order to show the extent to which am¬ 
biguity may arise, an odd number of panels has been selected 
in the present example. As has already been shown, the 
method of Art. 7 cannot be applied to a double system of 
triangulation. It is only possible to assume that each sys¬ 
tem acts as an independent truss, and to determine by trial 
the greatest possible concentrations at the head of the train, 
and in one system, and consider such concentrations the head 
of the moving load in that system. 

There will be two equal concentrations represented by w at 
the head of the moving load in each system, while those that 
follow will uniformly equal w. Ordinarily no two concentra¬ 
tions will be exactly equal, but the assumption is sufficiently 
accurate. 

The truss under consideration is composed of two systems 
of right-angled triangulation, shown in Figs. 3 and 4 of PI. II. 

Before passing to the computations it is well to observe 
that although the action of the loads in one system may be 
considered as taking place independently of the actions of 
the loads in the other; at the same time equal loads sym¬ 
metrically placed in reference to the centre, though resting on 
different systems of triangulation, may be considered counter¬ 
balanced. The web stresses due to the fixed load will be de¬ 
termined on the supposition that the web members shown by 
the dotted lines do not exist. 

The data to be used are given below: 

Span =210 feet. Depth of truss = 26 feet. 

Number of panels = 15 Panel length = 14 “ 

W (upper) = 9100 lbs. = 4.55 tons = 650 lbs. per foot. 

W' (lower) = 14000 “ = 7.00 “ = 1000 “ “ “ 

w — 13 tons. w' = 20 tons. 

e = w' — w — 7 tons. 


TWO SYSTEMS OF BRACING. 


6 1 


Angle QNO — ex Angle MNO = f 3 

tan a — 1.077 sec a = 1.47 

tan (3 =0.538 sec (3 =1.136 

The excess ^ will be taken at four panel points as before; it 
will also be used in determining the chord stresses. 

The counterbrace 16 is the first one needed. Carrying the 
moving load on the bridge from R, panel by panel, the great¬ 
est web stresses are found to be the following: 


In brace 1 6. w sec a + \\ e sec a — 

+ 

22.15 

tons. 

a 

11 

1 5 • • il w sec a + if e sec a — 

+ 

28.62 

a 

it 

ll 

14. .If w + ff e + W — 

— 

19.62 

11 

a 

ll 

1 • • (ff w + li e ) sec a + ( W + W ) sec a 

— + -52.06 

11 

ll 

n 

J 3 - -t! w + \\e + W = 

— 

24.02 

11 

ll 

ll 

2 • • (fl w + if e + W + W) sec ex. — 

+ 

59.80 

a 

n 

ll 

3 .w-h {^e + 2W + W' = 

— 

39-97 

ii 

it 

ll 

4. . {ff w + ff e + 2 ( W + W )} sec a 

+ 

84-53 

it 

it 

It 

3 . w + \\ e + 2W + W' — 

— 

45-23 

ti 

ll 

ll 

6 . . { ft w + t 4 e + 2 ( W + W') ( sec a 

+ 

93-54 

11 

It 

ll 

7. + ±%e + 3W+2W' = 

— 

62.05 

11 

li 

ll 

8..{ff w + \\e +3 (W + W')\ sec a 

= + 

H 9-53 

11 

(t 

ll 

9. .ff w + ff e + 3 W + 2 W' = 

— 

68.18 

u 

It 

ll 

10. . w + ffe + 3 (IV + W’)\ sec ft 

= + 

100.33 

11 


62 


SPECIAL NON-CONTTNUOUS TRUSSES. 


In brace 11. .w 4- W' — 4 - 27.00 tons. 

“ “ 12.. 7 (w 4 - W 4- W') see ft 4 - e sec ft 

= — 221.73 “ 

The stresses in each system of triangulation are found by 
virtually taking that system as a single truss supporting only 
the weights at the apices belonging to it. 

The greatest chord stresses will be obtained by supposing 
the train to cover the entire bridge, with the four excesses e 
at panel points 1, 2, 3, and 4. 

Greatest stress in a — 3 (w 4- W + W') [2 tan ft 4- (tan ft 
4- tan a)] 4- {w + W 4- W f ) tan ft + e (4 tan ft 4- tan a) 4- 
\e 2 tan ti — — 236.51 tons. 

Here it should be explained that since ff e — 3% e is found 
in the reaction at R, the three e’s at panel points 1, 2, and 3, 
and ^ of that at 4 may be taken as passing directly to R, 
while | of the e at 4 passes to M through 5', 2', 16, 14, 1, 3, 
etc. Counterbrace 16 thus comes into action. 

Greatest stress in b — [2 (w + W + W') + e] tan a 4- 

236.514 = — 291.91 tons. 

“ “ “ c = 2 (w 4- W + W') tan a + 

291.907 =-344.79 “ 

“ “ “ d = \_{w + W 4- W') — -§ e] tan a + 

344.787 = - 366.20 “ 

“ “ “ e = (w 4 - W 4 - W f ) tan a 4- 

366.201 = — 392.64 “ 

The panel stresses in e, f, and g will be the same; and if 
the loading were uniform over the whole bridge, the panels e, 
f, g, h, and k would all be subjected to the same stress. 

Greatest stress in / and m — [7 (w 4- W 4- W') 4 - 3^] tan ft 

— 4- 105.01 tons. 

“ “ “ n = [3 {w 4- W 4 - W') 4- i^e] tan ft 4- 

105.009 = 4- 149.65 “ 


TWO SYSTEMS OF BRACING. 


63 


Greatest stress in 0 — [3 (w 4- W 4- W') 4- e] tan a + 

149.654 = 4- 236.51 tons. 

“ “ “ / = [2 (w + W 4 - W) 4 - 1 *] tan a + 

236.513 = + 291.91 “ 

1 

“ “ “ q = 2 (w + W + JF') to a + 

291.906 = 4- 34479 “ 

“ “ “ r = (w -f W + W' — f e) tan a 4- 

344.786 = 4 - 361.17 “ 

u u “ s = (w + W 4- W ) ta7i a + 

361.174 = + 387.61 “ 

It is to be noticed that diagonally opposite panels in the 
upper and lower chords, up to the counterbrace 16, beginning 
with the panels a and 0, are subjected to the same amounts 
of stress, but of opposite kinds. 

If the loading were uniform over the whole bridge, this 
equality of the pairs would continue to the centre; also, the 
stresses in the panels e> f, g, h, k y and s would be equal to 
each other. 

If the end posts were vertical, there would be obvious 
changes in the stresses of the panels /, m } and n (that in l 
would be zero). The upper chord panel stresses would not 
be changed. 

The whole truss in Fig. 2, PI. II., is composed of the two 
systems of triangulation shown in Figs. 3 and 4, and each of 
these is to be considered separately in checking the chord 
stresses by the method of moments. Denote by (R r ) and 
(R") the reactions at the points indicated by the same letters 
in Figs. 3 and 4, then divide the total load supported by 
each system into two parts, according to the principle of the 
lever, and there will result: 

(R') = [7 (w + W+ W')] 4-• 2 e = 103.7867 tons. 

(R") — ^ [7 (w + W+ fV')] -f ff • 2 e — 91.3966 tons. 


64 


SPECIAL NON-CONTINUOUS TRUSSES. 


If the diagonals are in tension, according to this value of 
(R '), D'L should be drawn, and not K'E'. The latter is 
taken, however, for a reason that will appear presently. 

The sum of {R) and (R") is just equal to the total reaction 
at R in Fig. 2, as it ought to be. 

Indicate by ( BC ), (. D C '), (d), etc., the stresses in the panels 
represented by those letters. Taking moments about H and 
G respectively, there result: 

(BC) = -(HK) = [(R’) xR'H-2(w + W+ W')(GH +\FG)~]+d. 
(AB) = -(GH)=[(R)xR'G-(w' + W+ W')FG]+d. 

Also, taking moments about K' and H': 

(C , E>') = [(R n )xR"IC-2(w'+ JV+ W’) (H'K’ + iG'H')]+d. 

\B’C') = -(H'K') = [(R")xR' f H'--(w'A W+ W) G'H') + d. 

Similar expressions will give the chord stress in every panel 
of Figs. 3 and 4; and having found these, the resultant 
stresses in Fig. 2 are simply the sums of the proper pairs 
taken from Figs. 3 and 4. 

Thus, {d) - ( BC ) + {C'D') 

M = (CD) + (C'D') 

(0) = (GH) + {G'H') 

etc. = etc. + etc. 

This system of determination by moments may be applied 
to any truss with parallel chords, however many systems of 
triangulation there may be. 

The method also applies to any irregular loading, for the 
stresses due to each panel load may be found separately, and 
the sum caused by all taken. 

Web stresses may also be checked by the same method, 
since the increment of chord stress at any panel point is equal 
to the sum of the horizontal components of the stresses in 
the web members intersecting at the panel point in question. 
Such a check, however, is a very tedious one. 

Applying the above equations to C'D' in Fig. 4 : 

(C'D') = (91.397x84 - 2 x 31.55 x 42) -s- 26 = 193.35 tons. 


TWO SYSTEMS OF BRACING. 


6$ 


Also, to CD in Fig. 3 : 

(CD) = (103787 x 98 - 2 x 31.55 x 70 — 24.55 x 28)^26 = 

194.95 tons. 

But the sum of these two is 388.3 tons, whereas ( e ) — 392.64 
tons. This discrepancy, not very great, is easily explained. 
The loading (w + W 4- W') is counterbalanced in Fig. 2, but 
is not in Figs. 3 and 4. 

In Fig. 2 all the load on the left of the centre of the span, 
except \e at 4 or H', is assumed to pass directly to R (or 
R' and R"). Hence in Figs. 3 and 4, to be consistent with 
Fig. 2, there should be taken : 

(R) = 4 (w + W + W) + 2 e = 112.2 tons. 

(R ) = 3 (w + W + W) + fe = 82.983 tons. 

Introducing these in the general formula: 

(e) = (CD’) + (CD) = [112.2 x 98 + 82.983 x 84 — 31.55 x 
84 - 31.55 x 140 - 24.55 x 28] 4 - 26 = 392.75 tons. 

This result agrees sufficiently well with that obtained by 
the trigonometrical method. 

With the last value of (R"), ICE’ will be in tension. 

It is thus seen that with an uneven number of panels a 
little ambiguity exists both in reference to the greatest chord 
stresses and the greatest web stresses, when there are two 
systems of triangulation. This ambiguity always exists , 7what¬ 
ever the number of systems , if the component systems are not 
symmetrical in reference to the centre line of the span , and it 
always disappears if they are symmetrical in reference to that 
line. 

With an even number of panels in the span and two sys¬ 
tems of triangulation no ambiguity exists. 

These observations in reference to ambiguity apply as well 
to isosceles bracing as to vertical and diagonal. 

In the example taken there are only two systems of triangu¬ 
lation, but precisely the same method is to be followed what- 


66 


SPECIAL NON-CON TIN UO US TRUSSES. 


ever the number; in determining the web stresses, each sys¬ 
tem is supposed to carry those moving weights only which 
rest at its apices, and the same is true in reference to chord 
stresses for unsymmetrical loading, uniform loading being 
supposed counterbalanced for either stresses. 

The slight changes to be made for an overhead bridge, or 
for verticals in tension and diagonals in compression, are evi¬ 
dent from what has already been given in preceding articles. 

It is seen that any two web members intersecting in the chord 
not traversed by the moving load receive their greatest stresses 
at the same time; the principle, indeed, is a general one. 

When built in iron, this truss is frequently called the Lin- 
ville truss. 


Art. 13 .—Truss with Uniform Diagonal Bracing— Two Systems of Tri¬ 
angulation. 

This truss is shown in PI. X., Fig. 6, and, although taken 
here as an ordinary pin connection bridge, precisely the 
same method of calculation is to be used for a “ lattice ” truss 
with riveted connections. 

No locomotive excess will be taken, but a heavy moving 
load of uniform density will be assumed. The following are 
the data: 

Span = 182 feet. Depth = 23 feet. 

Panel length — 13 “ Number of panels = 14. 


Fixed load: 

W (upper) = 450 pounds per foot = 2.925 tons per panel. 
W' (lower) = 800 “ “ “ = 5.2 “ “ “ 


Moving load : 

w — 2800 pounds per foot = 18.2 tons per panel. 
Angle AaB — a. 

tana =0.565. sec a = 1.15. 

Wsec a — 3.364 tons. W' sec a — 5.98 tons. 

w 


14 


sec a — 1.500 


W tan a = 2.94 


u 


u 


W tan a 
w tan a 


1.653 “ 
10.28 




TRUSS WITH UNIFORM DIAGONAL BRACING. 


67 


The vertical members aB and tS are for tension only. 

The moving load will be taken as passing from A to T , and 
its head will be supposed to rest at the various panel points 
in succession, in the determination of the web stresses. 

The notation for the stresses is one which will frequently 
be used hereafter. The stress in any member is indicated by 
inclosing in a parenthesis the letters which belong to it in 
the figure. 


Head of moving load at D. 

(( dF ) = {J W 4- Wsec ol— 4- 2.6 x sec a . 

Hence the stress in dF will always be tension. 

Head of moving load at E. 

(Be) = “ W' — | W\ sec a — — 1.79 x sec a. 

Hence the stress in Ee will always be compression. 
The web stresses desired are, then, the following: 


(Ff) = 

-(iW' 

+ 

W - T \ zv) 

sec a 

= 4 - 

7.15 tons, 

i*G) = 

( W' 

+ 

\ W — t 6 4 ™) 

u 

=r — 

i -34 

u 

(Gg) = 

-( 


i W — \\ w) 

u 

= + 

16.32 

u 

(/#) = 

( w 


- ~h™) 

lc 

= — 

10.51 

u 

(HA) = 

( w 


4 - If zv) 

u 

= + 

26.99 

u 

(gK) = 

-( 


i W 4 - U zv) 

u 

— — 

I9.68 

<< 

(Kk) = 

( W' 

+ 

h W 4- \ f- zv) 

(t 

= + 

37.66 

u 

ihL) = 

-( w 

+ 

iv + a zv) 

(( 

= — 

30-35 

u 

(LI) = 

(I w 

+ 

IV 4 - f f zv) 

u 

= + 

49-83 

u 

(W) = 

- ( W' 

+ 

iiW+f%zv) 

u 

= — 

41.02 

u 

(Oo) = 

( 2W' 

+ 

H W + yf zv) 

u 

= 4 - 

62.00 


m = 

-(\\W' 

+ 

2 W + ff zv) 

u 

— — 

53.20 

u 

(Pp) = 

(2j IV' 

+ 

2 W + yi ZV) 

u 

= 4 - 

75.68 

u 

(oQ) = 

- ( 2 W' 

+ 2 IW + f%w) 

u 

— — 

65-37 

u 

m = 

(iW' 

+ 2%W+ f}zv) 

u 

= 4 - 

89-35 

u 

(PS) = 

- (2j W' 

+ 

3 W 4- -f £ zv) 

u 

= — 

79.04 

u 

(tT) = 

-±HW' 

+ 

W + zv) 

u 

— — 

197.24 

u 

(tS) = 

3 W 

+ 

3W 4- ffzv 


= 4 - 

90.68 

u 


68 


SPECIAL NON-CON TIN UO US TRUSSES. 


With the moving load covering the whole bridge, the fol¬ 
lowing chord stresses are found: 


(ad) 

— 

— (9J W ' + 9 W + 9J w ) 

tan a 

— — 141.46 

tons. 

(be) 

— 

( ad ) -5 ( W ’+ W + w ) 

u 

= - 215.83 

u 

(cd) 

— 

(be) -4( “ “ “ 

) 

u 

= - 275-33 

n 

(de) 

— 

(cd) -3( “ “ “ 

) 

it 

= - 3 > 9-95 

u 

W ) 

— 

(de) - 2 ( “ “ “ 

) 

u 

= - 349-70 

u 

(fg) 

— 

s 

1 

) 

u 

= - 364-57 

u 

(AB) 

— 

6 h ( “ “ “ 

) 

a 

= + 96.68 

u 

(BC) 

— 

(AB) +(2ifT' + 3 fP 4 -2j 

w ) 

u 

= + 134-69 

ll 

(CD) 

— 

(BC) + t >( W ' + W + w ) 

u 

= + 209.06 

u 

(DE) 

— 

(CD) + 4( “ “ “ 

) 

11 

= 4- 268.55 

u 

(EE) 

— 

(DE) + 3( “ “ “ 

) 

u 

= + 3 I 3 -I 7 

u 

(EG) 

— 

(EE) + 2( “ “ “ 

) 

u 

= + 342.92 

a 

(GH) 

— 

(EG) + ( “ “ “ 

) 

u 

= 4 357 -So 

<< 


The following operations constitute a check on the accuracy 
of the chord stresses. 

The horizontal forces exerted at the joints g and H, respec¬ 
tively, are: 

(fg) — \ Wtan a = — 365.40 tons, 

and ( 677 ) + J ( W' + w) tan a = + 364.41 “ 


The horizontal force exerted at either one of these joints, 
as found by the moment method, is: 


7 (W’ + W + w) x 0.25 x 182 

23 


= 364.5 tons. 


The agreement is close. 

It is to be observed that (Ff) is the greatest tensile stress 
in hL , also; and, on the other hand, that (kL) is the greatest 
compression stress Ff. Similar observations apply to the 
pairs of members eG, kK; Gg , Kg; fH, Hh. 

These, consequently, are the only web members which need 
to be counterbraced. 

Precisely the same methods of calculation apply, whatever 



COMPOUND TRIANGULAR TRUSS. 


69 

may be the number of systems of triangulation or the char¬ 
acter of the load, or whether the truss be a through or deck 
one. 

If Fig. 6 represented a deck truss, however, the compres¬ 
sive web stresses would be increased and the tensile ones 
diminished, while the chord stresses would remain the same. 
Since the increase of compression in any web member would 
numerically exceed the decrease of tension in the adjacent 
one, the truss is better adapted to a through load than a 
deck load. 

This truss, particularly with only one system of triangula¬ 
tion, is frequently called the “triangular” truss. 

Art. 14 .—Compound Triangular Truss. 

A very economical style of truss, in point of quantity of 
material, is that shown in Fig. 1 of PI. III. The truss is of 
the ordinary isosceles bracing, and formed of two systems of 
triangulation, but a half of the floor system and moving load 
is carried by verticals directly to the intersections E , F , etc. 

Half the weight of the trusses is supported at the apices of 
the main systems, as H and J/, in the upper chord, and half 
at the apices, as P and R , in the lower chord. The truss 
chosen is a deck or overhead truss; consequently half the 
floor system and moving load will be supported by the verti¬ 
cals in compression. The weight of the floor system will be 
taken at 300 pounds per foot, and the moving load taken will 
be a uniform one made up of a load of heavy engines weigh¬ 
ing 2700 pounds per foot. In such a case there is no ex¬ 
cess e. 

The following are the data: 

Length of span = 200 ft. Depth of truss = 27.75 ft. 

Upper-panel length = 12.5 “ tan CDL = tan a = 0.9 

Lower-panel length = 25.0 “ sec CDL —sec oc = 1.345 

W (upper) =25 x 500 + 12J x 300= 8.125 tons. 

W x (middle) = \2\ x 300 1.875 “ 

W' (lower) =25 x 500 = 6.25 “ 

W—w' — \ 2 h X 2700 = 16.875 “ 


7 ° 


SPECIAL NON-CON7INUOUS TRUSSES. 


The middle loads W 1 or w are applied to the trussing as 
follows. The adjoining figure represents a portion of the 
truss in question as indicated by the same letters HMPR (see 

figure in plate). Any weight rest¬ 
ing at K is carried down to the in¬ 
tersection, or two apices A and B, 
and the proper portion of each 
load is hung at each apex. In the 
truss in question, A C, in the ad¬ 
joining figure, will represent of 
the weight at K , and BD T 5 -g- of the 
same weight. The moving load is supposed to pass on the 
bridge from A. By examination it is seen that o and j are 
the first members which need counterbracing. The head of 
the train must be at the panel point between f and e for 
greatest moving-load stress in s , and at the panel point be¬ 
tween f and g for that in o, and at corresponding positions 
for other web members. 

Fixed-load stress in s .. = i( W + W t ) sec a = — 6.73 tons. 



ZV 

Moving-load stress in s. . = (1 +3 + 8 + 5) — sec a 

3 2 

= + 12.05 “ 

Fixed-load stress in 0 .. = |( W' + W t ) sec a — + 5.4 6 “ 


w 


Moving-load stress in 0.. = (1 + 4 + 3+ 5 + 12) —sec a 

32 


Fixed-load stress in p.. — \ W sec a 


= “ 1773 “ 

= - 546 “ 


w 


Moving-load stress in p.. = (1 + 3 + 8 + 5 + 7) —sec a 

32 

= + 17.02 “ 


Fixed-load stress in r .. = J W sec a 


— + 4.20 


u 






COMPOUND TRIANGULAR TRUSS. 




Moving-load stress in r .. = (i + 4 + 3 + 5 + 12 + 7) — sec a 

3 2 

= — 22.70 tons. 


Greatest stress in 


a 


ti 


u 


u 


it it 


ll ll 


1 C a u 


ll ll u 


Cl n 


ll ll ll 


ll ll ll 


ll It ll 


ll ll 


ll ll ll 


It ll It 


It ll ll 


1 .. = (|| w + \ W) sec a 

= - 33-82 “ 

2 .(|| w + | W' + | W i) sec a 

= + 34-53 “ 

3.. = (||w + iW 7 ')^ a 

— + 26.9 “ 

4 • • — (If w 4 - \ W + | « 

= ~ 41-47 “ 

5 • • = (ft w + i(W'+ W x )+ W) sec a 

= — 59.64 tons. 

8.. = (|f |(fF'+2^) + j IV) sec a 

= — 68.72 tons. 

• 

6 .. = (|| w+ iW+ W+iW,) sec a 

= + 49.87 tons. 

7 --=(U*> + hlV+ W r + IV,) sec a 

= + 58.93 tons. 

9.. = (ltw + JJF + W 7 ' + ^ 1 ) ^ ar 

= — 86.88 tons. 

12.. = (|J zu + \ W + W’ + | ^ or 

— “ 97.38 tons. 

11 • • = (Jf w + f W’ 4- J>Fi + ^ « 

— + 77* 1 3 tons. 

10 .. = (f |«; + | -f | W, + W) sec a 

— + 87.59 tons. 

1 3.. = (-W- w + f + f PFj 4- 2 W) sec a 

= —118.37 tons. 


SPECIAL NON-CONTINUOUS TRUSSES. 


72 

Greatest stress in 


« 


a 


u 


u 


a u 


a a 


u (( 


u u 


16 .. = (yy 8 - W + I W' 4 - 2 W x + 2 W) sec a 

= — 130.30 tons. 

l 5 • • = + i W + 2W' + % Wi) sec a 

= + 105.79 tons * 

14.. = (-y/- w + jJV + 2W' + 2 WJ sec a 

= + 117.66 tons. 

17 .. = yy 8 - 4- 2 w+ 2W + 2W X 


= —100.00 


« 


18.. = W + w 


— 25.00 “ 


The stress in 17 added to the vertical component of the 
stress in 16 is equal to (8w 4- 4W 4- 4^ 4- 3J W) the weight 
of the truss and its load, as it should. This constitutes a 
check in the work if, as was done, each web stress is found 
by adding a proper increment to a preceding one. 

For the greatest chord stresses the load will cover the whole 
bridge. 

Stress in a or b .. = (3J w 4- %W 4- 2W 4 - 2 W t ) tan a 

= — 78.75 tons. 

“ “ c or d. . — (6 w + 3 W' 4- 3 W x 4 - 3 W) tan a 4- 

78.75 =-213.75 “ 


u « 


a 


e or f.. = 2 (2 w + W + U\ + W) tan a + 

213-75 = - 30375 “ 

“ g or h. .= (2 w + W + W,+ W’) tan a + 

30375 = - 34875 “ 


a a 


a a 


n 


m 


“ “ / 


= (4W 4- 2 IT 4 - 2W 1 + £ W') tan a 

= 4 - 87.19 

= 3 (2w 4- W 4 - W x + W') tan a 4- 

87.19= + 222.19 

= 2 (2ie/ + W + W l 4- W') tan a 4- 

222.19 — + 312.19 “ 


a 


a 


COMPOUND TRIANGULAR TRUSS. 


73 


Stress in k. . = (2 zv + W + W t + W') tan a + 

312.19 = + 357.19 tons. 


In determining these values, it is to be remembered that 
the increment of chord stress at any panel point is equal to 
the sum of the horizontal components of the stresses in the 
web members intersecting at that point. 

The results for g or h or k may be easily verified by the 
method of moments. Let / be the span in feet, and d the 
depth of a flanged beam, in feet also ; then if w is the load 
per foot, the flange stress at the centre, as is well known, will 

•wl 2 

be —7-. To apply this to the present case, {w + W+ Wj + W') 
od 


must be written for w, and / and d have the values respect- 

iv 

ively of 200.00 and 27.75. Hence — 360.4 tons. 

od 


Now since the resultant stress at either of the centre joints 
is horizontal in direction for a uniform load from end to end 
of the truss, the value corresponding to the above will be 
found by adding to the stress in h the horizontal component 
of the stress in brace 1, for the supposed uniform load ; or by 
adding to that in panel k the horizontal component of that in 
brace 3. 


Horizontal component in 


( w + 


tan a — 11.25 tons, 


and 


348.75 + 11.25 = 360.00 tons. 


Horizontal component in 3 = 


W' 

2 


tan a — 2.8125 tons, 


and 357- x 9 + 2.81 = 360.00 tons. 

Both of the above results are remarkably satisfactory verifi¬ 
cations ; they would have agreed exactly, but 0.9 is not the 
exact value of tan a. 

If the bridge were a through one, the general method of 






SPECIAL NON - CON TIN UO US TRUSSES . 


74 

calculation would be exactly the same; the slight changes in 
the details of the operations are sufficiently obvious after 
what has been said before. 

As a through truss there would be some saving of material, 
for the secondary verticals 18 would be in tension. 

A much greater saving might be effected by using inclined 
end posts, in which case a short beam or girder would take 
the place of the end panels LC , and braces 15 would be verti¬ 
cal and run up to Z, while braces 18, 14, and 17 would be 
omitted altogether. 


Art. 15.—Methods ot Obtaining Stresses—Stress Sheets. 

In the preceding cases the analytical expressions for the 
stresses have been written in such a manner as would seem 
best to show in detail the principles by which they are 
traced. 

In practice every engineer has a method best fitted for 
himself by either habit or taste. 

The “ strain sheet,” or properly “ stress sheet,” is almost in¬ 
variably made as shown in the plates. A skeleton of the 
truss is drawn, and along each member is written the great¬ 
est stress belonging to it. 

Art. 16.—Ambiguity caused by Counterbraces. 

It is important to notice, from what has preceded, that a 
little ambiguity always exists, both in web and chord stresses, 
near the middle of the truss, when counterbraces are used in¬ 
stead of counterbracing. This arises from the fact that even 
with a single system of triangulation it is impossible to divide 
the truss by cutting less than four members, which is equiva¬ 
lent, as a question of equilibrium, to having four unknown 
quantities and only three equations by which they are to be 
determined. 

This ambiguity, however, has been shown to be not of a 
dangerous character. 


CHAPTER III. 


NON-CONTINUOUS TRUSSES WITH CHORDS NOT PARALLEL. 


Art. 17.—General Methods. 


The determination of stresses in trusses with chords that 
are not parallel can usually be accomplished more con¬ 
veniently by either a combination of the method of moments 
with the graphical method, or the graphical method alone, 
than in any other manner. 

So long as three members, at most, are cut by any surface 
whatever, dividing a truss into two parts, the problem of the 
determination of the stresses in these members is determin¬ 
ate; for in such a case the problem is really one of the equi¬ 
librium of any system of three forces parallel to a given 
plane, for the solution of which, as is well known, there are 
three equations of condition. 

The matter, however, requires a little attention here, in 
order that the particular kind of stress (tension or com¬ 
pression) developed in 



any bar may be known 
from the stress dia¬ 
gram. 



Let Fig. i represent 
a portion of any truss 
divided into two parts 
by the plane (it might 
be any other surface) 
AB ; let F , G, and H 
be the points of inter, 
section of this plane 
and the three mem¬ 


L 


Fig. i. 


bers CK, CD , and ED; and let 2 P be the resultant of all the 
external forces acting on that portion of the truss lying on 


7 5 







76 


NON-CON TIN UO US TRUSSES IN GENERAL. 




c 

Fig. 2. 


the left of AB. The external forces are known, and the stresses 
c, t\ and t, in CK, CD, and ED respectively, are required. 

Now, any one of those stresses may be determined by the 
method of moments, if the origin of moments be properly 

located. If the origin be taken at the 
point of intersection of the lines of ac¬ 
tion of any tzvo of the stresses , the 
moments of those stresses will be zero. 

Hence, as a general principle, in 
order to determine any one of those 
unknown stresses by the method of mo¬ 
ments, the origin of moments is to be 
taken at the intersection of the lines of 
action of the other tzvo ; the moment of the third unknown 
stress can then be placed equal to the resultant moment of 
the external forces, giving one equation with one unknown 
quantity. 

Suppose in Fig. i that the stress in CK is to be found by 
moments. D is the point of intersection of CD and ED, and 
consequently is the origin of moments. The lever arm of 
that stress is of course the normal distance from D to CK. 
The kind of stress in CK can always be determined by the 
known direction of the resultant external moment. If the 
effect of the moment is to shorten the piece in question, the 
resulting stress will be compression, while tension will exist if 
the effect is to lengthen the piece. 

If any one of the three stresses is known, either of the 
others may be found by moments, by taking the origin of 
moments anywhere on the line of action of the third force. 
This method is illustrated on page 255. 

In Fig. 1 it will be assumed that the effect of the moment 
of the external forces on the left of AB is a tendency to turn 
that portion of the truss in the direction of the curved arrow, 
and consequently to shorten CK, thus producing compression 
in that member. Let c represent that compression, then, so 
far as the portion of the truss on the left of AB is concerned, 
it is equivalent to an external force acting from F toward C, 
as shown in the figure by the arrow c. 





GENERAL METHODS. 


77 


Now, it is known from rational mechanics that if the sides 
of a closed polygon represent a system of forces in equilib¬ 
rium, the lines representing the forces must be laid off in the 
same direction around the polygon, in order that the true 
direction of those forces may be represented. If, for instance, 
the lines of the polygon in Fig. 2 represent a system of forces 
in equilibrium, the forces must act in the direction shown 
by the arrow-heads : all in the same direction around the 
polygon. 

The portion of the truss on the left of AB, Fig. 1, is held 
in equilibrium by the external forces acting upon it, and the 
three stresses acting at F, G, and H. The amount and direc¬ 
tion of the stress acting at F have already been found ; they 
are represented by c. 

In Fig. 2 let 2 P represent the resultant external force act¬ 
ing on the left of AB, both in amount and direction, as shown 
by the arrow-head ; then draw c parallel to CK, representing 
in amount and direction of arrow-head the compression in 
CK; finally, complete the quadrilateral by drawing t and t' 
parallel to ED and CD respectively. The directions of c and 
2 P are fixed already, hence t and /' must have the directions 
indicated by the arrow-heads in those lines. The directions 
and magnitudes of the forces t and f are thus known, and H 
and G are their points of application. Finally, consider t and 
t' attached to their points of application, as indicated in Fig. 

I ; their tendencies will be to lengthen EH and CG respect¬ 
ively, hence ED and CD will both be in tension. 

The general method, then, of determining the kind of stress 
existing in any member, is to apply to the point of division of 
that member the force represented in magnitude and direction by 
the proper side of the equilibrium polygon, and observe zvhether 
the effect is to shorten or lengthen the member in question: in 
the first case, the stress will be compression ; in the second, ten¬ 
sion. 

This method is perfectly general, and may always be used. 

The origin of moments for the stress in CD would be the 
intersection of CK and ED produced, and C would be the 
origin for the stress in ED. 


78 


NON-CON TIN UO US TRUSSES IN GENERAL. 


So far as the method is concerned, it is a matter of indif¬ 
ference which one of the three stresses is first found by mo¬ 
ments ; it is simply advisable to take that one which can be 
found most conveniently. 

In all trusses with vertical loading, 2 P is the algebraic sum 
of the external forces acting on the portion of the truss in 
question, or the “ shearing stress .” 


Art. 18.—Curved Upper Chord—Two Systems of Vertical and Diagonal 

Bracing. 

The preceding principles will be first applied to the truss 
shown in Fig. 2, PL III. That truss is taken first, rather than 
a simpler one, because the application of the method is there 
perfectly general, including all cases possible. 

The truss taken is composed of two systems of triangulation, 
and it is of the greatest importance to notice that the systems 
cannot be treated separately, since the upper chord is curved, 
and the loads supported by one system induce stresses in the 
other. As an extension of this part of the matter, it is equally 
true that with any number of systems of triangulation, of 
whatever kind, any load on one system will induce stresses in 
all the others. The result of this is, that with a curved chord 
and more than one system of triangulation there is always 
ambiguity in the determination of stresses, because more than 
three members will be severed however the truss may be 
divided. Certain assumptions, however, may be made which 
involve no danger, and which give a determinate character to 
the stresses desired. 

The assumption in general terms is this— i. e., that at the 
point for which 2 P is zero the total load is divided, and each 
of its portions travels to the nearest abutment by the most 
direct path. This is a legitimate analysis, so far as simple 
equilibrium is concerned, but it is by no means certain that 
the stresses determined are those which actually exist in the 
truss. 

In Fig. 2 of PI. III., if a uniform load covers the entire 
truss it will be assumed that the counters 18', 18, 19, 20, and 


CURVED UPPER CHORD. 


79 


22, and the corresponding ones on the other side of the 
centre, are not needed, and consequently not subjected to 
stress. 

Again, suppose the left half of the truss to be loaded with 
the moving load, in addition to the fixed load over the entire 
bridge, and suppose 2 P = o for the panel point at the foot 
of diagonal io; then it would be assumed that all the verti¬ 
cals are necessary, but only the diagonals 19, 20, 22, 24, 26, 
28, 30, 32, 34, 36, and 37 in one part of the truss, and 10, 8, 
6, 3, and 2 in the other part. 

These assumptions involve no danger, because the stresses 
deduced by one legitimate method of analysis at least are 
provided for. Nevertheless there is some ambiguity, and 
when that exists there cannot be economy of material. 

The greatest stresses to be determined in the case treated 
will be found in accordance with the preceding principles. 

The moving load will consist of a train headed by four 
excesses e , and will be supposed brought on from left to 
right, first touching the truss at R . 

The following are the data: 


Length of span = 200 ft. Height at centre = 35 ft. 

Length of panel = 12.5 “ Height at ends = 15 “ 

Number of panels = 16 Radius of upper chord = 260 “ 


W (upper) = 500 lbs. per foot = 3.125 tons 
W' (lower) = 800 “ “ “ =5.000 “ 

w — 13.00 tons. w' — 17.00 tons. 


_Fixed load. 


e — w r — w ~ 4.00 tons. 


The truss is supposed to be designed for a single-track 
through railroad bridge. 

The stresses due to the fixed and moving loads will be 
found separately, and those due to the fixed load will be found 
first. 

For any given condition of loading, the reactions at the 
two ends R and R' will be denoted by these letters simply. 

As the notation indicates, a part of the fixed load is assigned 


8o 


NON-CON TIN UO US TRUSSES IN GENERAL. 


to the upper chord, according to the principles already set 
forth. 

For the fixed load only, R = R' = 60.9375 tons = 7.5 
(JV + IV'). It was shown above that the verticals and main 
diagonals are the only web members which will be assumed 
to be stressed by the fixed load. 

Fig. 1, PI. IV., is the only diagram necessary for the deter¬ 
mination of all the fixed-load stresses in the truss. 

The truss may be divided through the members /z, 16, and 
k without severing more than three members; hence the 
stresses in those three are determinate. 

The stress in h is the most convenient one to find by the 
method of moments, hence the middle panel point of the 
lower chord at the intersection of k and 16 will be the origin 
of moments. The lever-arm of the stress in h is found by 
careful measurements (it might be found by calculation) on a 
large drawing to be 34.95 feet. Taking moments, therefore: 

(7.5 (IV + W') x 100 — y(JV + W') x 50)-f-34.95 — 93.1 tons = 

stress in /z. 

Hereafter, for the sake of brevity, the stress in any member 
will be represented according to the notation of Art. 13. 

Now let a dividing plane cut the truss in /z, 16, and k\ that 
portion on the left of it will be held in equilibrium by the 
external forces 2 P = 7 .%(W 4- W') — y(W+ JV') = 4.0625 
tons and the stresses (/z), 06 ), and (k ); their directions being 
determined according to the preceding principles. 

In Fig. 1, therefore, of PI. IV., make LM, acting upward, 
equal to 4.0625 tons, and /z, acting toward M, equal to 93.1 
tons; then draw k and 16 parallel to the members indicated 
by these letters; the directions of action of the stresses are 
indicated by the arrow-heads, being the same around the 
quadrilateral in question. Stresses (16) and (k) are therefore 
tension. 

The actual stresses were determined with a scale of ten 
tons to the inch ; but Fig. 1 of PI. IV. is drawn to a scale of 
twenty tons to the inch. 

The next plane of division cuts^, 16, 15, and £, but (k) and 


CURVED UPPER CHORD. 


8l 


(i6) are known, hence (15) and {k) may be determined. For 
this plane, Z2P = 7.$ (W + W) — 7 W’ — 6 W = \ W + \W— 
7.1875 tons. Lay off from L downward a distance (not let¬ 
tered) equal to 7.1875, and from the lower extremity of 16 
draw^* parallel to that panel of the upper chord until it inter¬ 
sects LM in 0 . As shown, (g) will act toward 0 , and the 
difference between LO and 7.1875 will be stress (15); it will 
act up, and consequently will represent tension; its value is 
1.9 tons, g, of course, represents compression. 

The third plane of division cuts g, 16, 14, and/. 2 P = 
j(W + W) = 12.1875 tons. Hence make OK equal to 
12.1875, aR d draw 14 and / parallel to those members; their 
directions are indicated by the arrow-heads. 

The completion of the diagram is simply a repetition of 
these operations; it is only necessary to use care in giving 
the stresses their proper directions. The stresses in the 
verticals will be found in the vertical line A T ; the shearing 
stresses 2P are also laid off on that line, acting upward. 

The following are the results of the complete operation, 
4- denoting tension, and — compression: 


(ti) = — 03.1 tons. 

( k ) = + 91.5 “ 

U) = - 93-25 “ 

(/) = + 89.2 “ 

(/)=- 92 -i “ 

(tn) = + 83.7 “ 

(e) = - 90.4 “ 

(n) = + 78.0 “ 

\d) = - 85.8 “ 

( 0) = + 66.0 “ 

(c) = — 80.5 “ 

(/) = + 49-7 “ 

(/>) = — 69.6 “ 

(,j) = + 20.0 “ 

(a) = - 52.7 “ 

(r) =0 

(1) = - 60.9375 


(16) = 

+ 

2.6 

tons. 

(* 7 ) = 

+ 

1.0 

u 

(i 5 ) = 

+ 

1.9 

u 

(14) = 

+ 

4.0 

a 

03 ) = 

— 

i -5 

u 

(12) = 

+ 

8.5 

u 

(1,) = 

— 

1.4 

u 

(10) = 

+ 

8.6 

a 

( 9 ) = 

— 

6-3 

u 

(8) = 

+ 

16.5 

u 

( 7 ) = 

— 

8.3 

n 

(6) = 

+ 

21.0 


( 5 ) = 

— 

I 3 - 1 

<< 

( 4 ) = 

— 

20.2 

a 

( 3 ) = 

+ 

bo 

u 

(2) = 

+ 

31.8 

it 


3.125 = - 64.0625 “ 


82 


NON-CONTINUOUS TRUSSES IN GENERAL. 


If the diagram has been properly constructed, the sum of 
the vertical components of (2), (3), and ( a ) will be 7.5 (W -f 
W') = 60.9375 tons. In this case it came 60.25 tons, which 
is sufficiently near to prove the accuracy of the work. Con¬ 
stant checks may also be applied in the course of the work, 
for the vertical component of the stress in any diagonal must 
be equal to the algebraic sum of the stress in the vertical 
which meets it at its foot and the weight which hangs from 
the same point. 

With the exception of a few web members near the centre 
of the truss, the greatest web stresses will exist at the head of 
the train when it covers the larger segment of the span, pre¬ 
cisely as in trusses with parallel chords. A number of the 
web stresses, however, which exist near the head of the train, 
for each of its positions, will be given for the main diagonals, 
since they appear in the stress diagram and may readily be 
scaled from it. 

Bringing the moving load on from R, panel by panel, it is 
found by trial that 18' is the first counter needed, the head of 
the train being at its foot. For this position of the moving 
load R' = 6.375 tons, hence the point for which 2 P = o is at 
the head of the train. It is then assumed that only the diago¬ 
nals 6, 3, and 2 on one side, and 18', 18, 19, 20, 22, 24, 26, 28, 
30, 32, 34, 36, and 37 on the other are needed. The truss 
may then be divided by a plane cutting the three members d, 
6, and p only, and any one of these may be found by the 
method of moments, but it is convenient to take d. The 
lever-arm of d is found to be, by a large scale, 26.7 feet. 
Hence, 

(d) = (R! x 162.5) ~=~ 26.7 = — 38.8 tons. 

A single four-sided stress polygon then gives: 

(18') = + 19.00 tons. 

The head of the train next rests at the foot of 18. The 
diagonals on the right of the head of the train which slope 
similarly to 18, and those on the left of it which slope simi- 


CURVED UPPER CHORD. 83 

larly to 8, are needed ; the others are not. R = 10.63, hence 
2 P= o at the head of the train. 

Hereafter the lever-arm of any upper chord panel will be 
denoted by /, with the proper subscript. In the present case 
4 = 29.7 feet; hence, 

(e) = R x 150 -r- 29.7 = — 53.7 tons, 

and a single diagram gives 

(18) = + 24.8 tons. 

In order to save constant repetition, it may be stated as a 
general rule that for every position of the train all the verti¬ 
cal web members are needed, but only those inclined braces 
which slope upward and away from that panel point for which 
2 P = o are considered necessary. 

With the head of the train at the foot of 19 ,R— 15.68 
tons, l f — 32.10 feet; hence, 

(f) = — 67.2 tons. 

Diagrams give 

(11) = + 4.0 tons. 

(19) = + 28.8 “ 

Only the diagrams for the next three positions of the mov¬ 
ing load will be given, for they will sufficiently illustrate the 
rest; they all embody exactly the same operations. 

Figs. 2, 3, and 4, of PI. IV., are all drawn to the same scale 
of 20 tons to the inch. 

With the head of the train at the foot of 20,^ = 21.56 
tons, and l f — 32.10 feet; hence, 

(/) = (R x 137.5 ~ W 'P) d- 3 2 - 10 = -85.75 tons, 

since 2 P = o at the foot of 19, and p is the panel length. 

Divfde the truss through f 19, and m, then in Fig. 2, PI. 
IV., make 1 — 6 vertical, and f parallel to that panel and 


8 4 


NON-CONTINUOUS TRUSSES IN GENERAL. 


equal to 85.75 tons; it acts toward 2 as shown. Make 2 — 3 
equal to 21.56 — w — 4.56 tons, and then draw m and 19. 
These latter must act in the direction shown. Now suppose 
g, 19, 13, and m severed. From the upper extremity of 19 
draw g parallel to that panel (referring to Fig. 2) until it in¬ 
tersects 1 — 6 in 1 ; 1 — 2, acting downward, is the tension 
in 13. 

It should be stated that the shear 2 — 3 = 4.56 tons acts 
downward, as shown. 

Next make 1 — 6— 21.56 tons, acting down, and draw / 
and 20, acting in the directions shown ; finally, draw h and//'. 
1 — 4, which acts upward (not indicated), is the compression 
in 15 ; and, in like manner, 4 — 5 is the compression in 17. 
Scaling from the diagram : 

(11) = 4- 5.2 tons. (20) = 4- 16.0 tons. 

( 13 ) = + 3*9 “ ( 15 ) = — 10.1 “ 

(17) = — 9.0 “ (19) =4- 17.6 “ 

With the head of the train at the foot of 22, R' — 28.25 
tons, and = o at the foot of 20; hence (g) is the stress to 
be found by moments. l g — 33.75 feet. 

(g) = ( R ' x I2 5 - W ’P) ^ 3375 = - 9 8 -3 tons. 

Fig. 4, PI. IV., is the diagram used for this position of the 
load, and is constructed in precisely the same manner as Fig. 
2. 2 — 5 is equal to R' — w' = 11.25 tons, the shear for^*, /, 

and 20, and it acts downward. 1 — 7 is the shearing stress, 
28.25 tons, for h, k, 22, and 20, also acting down, as shown. 
1 — 2 is the stress in 15 ; 2—3, that in 13 ; 1—4, that in 17; 
and 4 — 6, that in 23. Below are the numerical values: 

(15)=+ 3.8 tons. (13) = + 6.0 tons. 

(22) = 4 - 16.0 “ (23) — — 10.8 “ 

(20) =4- 20.8 “ i l 7)= —12.2 “ 

Fig. 3 is the diagram for the head of the train at the foot 
of 24, and is constructed in precisely the same manner as 


CURVED UPPER CHORD . 


85 

Figs. I, 2, or 4. R' = 35.75, and o for the foot of 20. 

4 = 3375 feet. Hence, 

U) = (R' x 125 - 30/'/)-*- 33.75 = - 113.5 tons. 

(15) = + 4-8 tons. (24) = + 25.2 tons. 

(22) = +15-0 “ (23) = - 9.4 “ 

(17) = - 3.2 “ • (25) =-157 “ 

With the head of the train at the foot of 26, ^P = o for 
the foot of 22, since R' = 35.75 tons. 4 = 34.65 feet. Hence, 

(A) = ( R' x 112.5 — 3 w’p) -f- 34.65 = — 124.7 tons. 

(22) = -f 16.1 tons. (26) = + 34.3 tons. 

(i 7 )=+ 7 -o “ (25) = - 2.5 “ 

(24) = +12.5 “ (27) = — 28.0 “ 

(23) = - 10.2 “ 


With the head of the train at the foot of 28, R’ = 53.1875 
tons, and 2 P = 0 at the foot of 22. l h = 34.65 feet. Hence, 


(k) = (R r x 112.5 — 6 w'p) -7- 34.65 = — 136.1 tons. 


(17) = 4- 6.9 tons. 

(24) = + 12.8 “ 

(23) = - 1.0 41 

(26) = + 21.5 “ 


(25) = — 1.5 tons. 

(28) = + 25.1 “ 

(27) = — 16.0 “ 

(29) = -15.5 “ 


With the head of the train at the foot of 30, R' = 63.125 
tons, and 2 P = o for the foot of 24. l h > = 34.95 feet. 


(//) = (R* x 100 — 6 w'p) + l h ,= — 144.1 tons. 


(17) = + 10.0 tons. 

(23) = + 4-5 “ 

(26) = + 15.7 “ 
(25) = + 1.0 “ 

(28) = + 20.8 “ 


(27) = — 9.3 tons. 

(30) = + 35-6 “ 

(29) — — 10.7 “ 

(.30 = — 27-5 “ 


With the head of the train at the foot of 32, R' = 73.875 
tons, and 2 P = o for the foot of 24. l h > = 34.95 feet. 


86 


NON-CONTINUOUS TRUSSES IN GENERAL. 


••• m = 


X IOO X 

10 w'p) 4' = 

— 150.6 tons. 

(2 7 ) = 

— 

9.9 

tons. 

(32) = 

+ 24.8 tons. 

(30) = 

+ 3 6 -o 

<< 

(31) = 

- 26.8 “ 

(29) = 

— 

1.0 

u 

(33) = 

— 16.7 “ 

With the head 

of 

the train at the foot of 34, P r = 854375 

tons, and 2 P 

-- O 

for 

the foot of 24. l h >; 

= 34.95 feet. 

• 

• 

• 

II 

X 100 - 

- 14 w'p — wp) -7- 34. 

95 = — 154.7 tons. 

(17)= 

+ 

8.8 

tons. 

(32) = 

+ 24.6 tons. 

(25) = 

+ 

8.9 

<< 

(31) = 

- 17.4 “ 

(23) = 

+ 

5-5 

a 

(34) = 

+ 55-2 “ 

(30) = 

+ 26.5 

u 

(33) = 

-15.4 “ 

(29) = 

— 

0.8 

a 

( 35 ) = 

- 46.8 “ 

With the head 

of 

the train at the foot of 36, R' — 97.8125 

tons, and 2 P- 

= 0 

for ' 

the foot of 24. 4' = 

: 34.95 feet. Hence, 

II 

>3 

X 

IOO — 

1 Zw'p 

- 3 wp) 34.95 = - 156.5 tons. 

(17) = 

+ I 

1.0 

tons. 

(32) = 

+ 12.0 tons. 

(24) = 

— 

2.0 

<« 

(31) = 

- 20.3 “ 

(23) = 

+ 

4-5 

u 

(34) = 

+ 60.0 “ 

(25) = 

+ 1 

2.0 

(t 

( 33 ) = 

- 2.8 “ 

(27) = 

— 

5.0 

u 

(36) = 

4- 40.0 “ 

(30) = 

+: 

59.8 

u 

(35) = 

-50.8 “ 

(29) = 

+ 

8.4 

u 



With the head 

of 

the train at the foot of 37, R' = 111.00 

tons, and ^P - 

= 0 

for 

the foot of 24. 4' : 

= 34 - 95 - Hence, 

X 

II 

IOO 

— 22 w’p • 

— 6 ivp) a- 34.95 = — 156.00 tons. 

(17) = 

+ 

10.9 

tons. 

(32) = 

+ 10.5 tons. 

(24) = 

— 

2.2 

<< 

(31) = 

— 16.7 “ 

(23) = 

+ 

4.4 

u 

(34) = 

+ 54-2 “ 

(25) = 

+ : 

11.9 

u 

(36) = 

+ 37 -o “ 

(28) = 

+ 

i -5 

u 

( 35 ) = 

- 43-5 “ 

(27) = 

— 

5-5 

<< 

(37) = 

+ 78.0 “ 

(30) = 

+ 26.0 

<« 

(38) = 

— III.O “ 

(29) = 

+ 

8.7 

u 




CURVED UPPER CHORD . 


8 7 


The chord stresses with this loading are: 

o 


( a ) = 

1 

00 

• 

tons. 

(*) = + 

157.6 tons. 

(*') = 

- 131.2 

u 

(0 = + 

150.5 

u 

( o ') = 

“ 137-0 

it 

( m ) = + 

149.7 

u 

(<*') = 

- 153-5 

it 

( n ) = + 

132.0 

it 

(/) = 

- 152.6 

it 

(*') = + 

124.5 

it 

(/') = 

- 158.7 

u 

(/) = + 

81.8 

it 

{£') = 

- 156.6 

(i 

(?) = + 

50.0 

It 

( A ') = 

- i 5 6 -o 

it 

(r) = 0 




The same checks apply as with the fixed load. The great¬ 
est stresses are found by combining the fixed and moving 
load stresses. The greatest chord stresses are thus found to 
be the following : 


(i a ') = — 140.4 tons. 
(£') = — 200.8 “ 
(V) = - 217.5 “ 
(d') = - 239.3 “ 

( e ) = - 243.0 “ 
(/') = — 250.8 “ 
W ) = - 249-85 “ 

(//') = — 249.1 “ 


V) = O 

(q') — + 70.0 tons. 

(V) = + 131-5 “ 

(o') = + 190.5 “ 

(n) = + 210.0 “ 

(m) = + 233.4 “ 

(/') = 4 - 239.7 « 

(k) = + 249.1 “ 


In the case of the web stresses the combination is effected 
by taking the algebraic sum. It will be seen that a number 
of the braces near the centre need counterbracing, i. e ., acting 
consistently with the assumptions that were made. It is 
uncertain to what extent the existence of the counters renders 
this counterbracing unnecessary; their influence is therefore 
neglected. 

It is to be borne in mind that the greatest result of a given 
sign is to be selected from all the preceding moving-load 
stresses, and added, algebraically, to the stress in the same 
member caused by the fixed load. 


88 


N0N-C0NT1NU0US TRUSSES IN GENERAL. 


(180 = 
(18) = 

+ 

+ 

19.0 

24.8 

tons. 

u 

(17) = • 

1; 

11.2 

12.0 

tons. 

a 

II II 

'Os'cT 

+ 

+ 

28.8 

20.8 

u 

u 

(23) = \ 

!; 

8.9 

74 

a 

u 

TT IT 

II II 

+ 

+ 

16.1 

27.8 

u 

u 

(25) = 1 

I; 

17.2 

10.5 

u 

a 

(26) = 

+ 

38.3 

u 

(2 7 ) = 

— 

29.4 

a 

(28) = 

+ 

33-6 

a 

( 20^ — *1 


21.8 

a 

(30) = 

+ 

44.6 

u 

\ z 9 ) ~ j 

[ + 

2.4 

u 

(32) = 

+ 

4 i -3 

u 

(31) = 

— 

35-8 

a 

(34) = 

+ 

8l.O 

u 

(33) = 

— 

29.8 

<< 

(36) = 

+ 

74.8 

u 

(35) = 

— 

71.0 

u 

(37) = 

+ 

IO9.8 

u 

(38) = 

— 

175.0625 tons, 


In actual practice it perhaps would hardly be worth while 
to counterbrace the web member 29. 

These results, then, are the greatest values of the stresses 
to which the different members of the truss are subjected. 


Art. 19.—General Considerations. 

It is clear that in the graphical treatment of such a problem 
the stress diagrams should be as large as possible. The scale 
used for all the results obtained in Art. 18 was ten tons to 
the inch, and it is not usually best to use more tons to the 
inch than that. 

Another method, but a far more tedious one, of applying 
precisely the same general principles is to determine the 
stresses produced in every member of the truss by each indi¬ 
vidual panel load, and then combine the results. The steps 
of the different operations in such a case are precisely the 
same as those gone through above. 

Although the truss taken consists of but two systems of 
triangulation, precisely the same method is applicable to any 
number of any kind of systems, or to the ordinary “bow¬ 
string’' truss of one system. 

In the latter case there is no ambiguity unless counter¬ 
braces are used. 


GENERAL CONSIDER A TIONS. 


89 


It is to be particularly noticed, also, that the method is per¬ 
fectly independent of the character of the curve of the upper 
chord ; it may equally well be applied to trusses with both 
chords curved, or to trusses with parallel chords ; in fact, it is 
perfectly general, though usually not desirable. 

Art. 20.—Position of Moving Load for Greatest Stress in any Web 

Member. 

Two principal cases occur in connection with types of 
structures ordinarily used in engineering practice. That one 
to be treated first is the case of the intersection of the chord 
sections, in any panel, lying below the inclined web member 
in the same panel; the other is the case of the intersection 
lying above the inclined web member. Applications of these 
principal cases to special features can easily be made after 
the general results are obtained. 


Case 1.—The Intersection of Chord Sections below the Inclined Web 

Member. 



Let l be the length of span; i the distance from end of 
span to the point of intersection, H } of the chord sections in 
the panel in question ; m the distance from the end of the 
span to the same panel, whose length is p; S the stress in 
the web member under consideration, and h its lever arm 
about H; a , b , c , etc., the distances separating W 1 from W 2i 
W 2 from W 3 , etc., etc.; W lt W r 2 , etc., the weights resting be¬ 
tween G and D, and W 3 , IV 4 , etc., the weights resting in the 
panel p, while W n , distant x from E t is the last weight rest- 

















90 


N0N-C0NTINU0US TRUSSES IN’ GENERAL. 


ing on the span from C to toward E. b' is the distance from 
D to the nearest weight, W 3 . 

The reaction at (T is : 


R 


/a + b 4- c + • • • + x\ 1jr (b + c 4- • • • + Tjrr — 

= WA — -^-y- J + etc. • • + W n ~ 


x 

1 


(i). 


By taking moments about H: 


Sh = Ri- W X {1 + i — a — b - x)-W 2 (/ + i-b-c - x) 

— etc. — ( w/- - - + w/- - ^ -- + etc.) ( m + i). . (2). 


P 


P 


By moving the entire load the distance Ax toward G, re¬ 
membering that the change in the value of R will be 

A X 

A R = (W 1 + W 2 + • • • + W n ) —j- , the change in Sh becomes: 


A Sh — ( W t + W 2 + • • • + lVn„) i — (W 1 + W 2 + etc.) Ax — 

(W 3 + W t + etc.)(m + 1) . . . (3). 

P 

For a maximum or minimum A Sh = o, hence : 

W\ + W 2 + • • • + Wu — ( W\ + IVt 5 + etc.) + (\ ■+■ etc.) 

l {in 4 - i) 

pi 

Bearing in mind that the first parenthesis in the second 
member of Eq. (4) represents the load between the panel p 
and the left end of the span, and that the second represents 
the load in panel p itself, it will be at once seen that when 
the load extends from E to W lf S is the maximum main 
stress; and that when it extends from G to W 4 (i. e., to the 
weight farthest toward E), S is the maximum counter stress. 
Eq. (4), therefore, as it stands, gives the condition for maxi¬ 
mum main or counter stress. 












GREA TEST WEB STRESSES. 


91 


Eq. (4) is perfectly general in character and covers all sys¬ 
tems of loading whatever, but it may be put in special forms 
for convenient application in special cases. 

Example I. — Uniform Load. 

If the load is continuous, or only partially so, and w is its 
intensity (u e ., its amount per lineal unit) at any point distant 
x from E, then the various concentrations W lf W 2 , W 3 , etc., 
will be represented by wdx. If, further, the load is uniform 
and continuous, w is constant, and W t + W 2 4- ... 4- W n = 
wx 1, x x representing the length of uniform load on the 
bridge. In the same manner, if x 2 represents the length of 
uniform load from D towards G,(W l -\- W 2 + etc.) = wx 2 ; and 
(W 3 + W 4 + etc.) = wrp\ r being the fractional part of the 
panel p covered by the uniform load w. 

Eq. (4) then becomes : 

/ / (m 4 - i) 

wx 1 = -. wx 2 4 - wrp -;—-. 

1 P l 



I . (ni \ 

x x = ~^x 2 4 r/h + I 1 . . . 



As with the general case so with Eq. (5), it is so written as 
to give both maxima, main and counter stresses. 

If the load is placed for the greatest main stress , for which 
x 2 = o, while X\ — up + rp\ in which np is the length of load¬ 
ing from C towards E : 

o 


np + rp ■=■ 



n 




If the load is placed for the greatest counter stress : 





9 2 


NON-CON TIN UO US TRUSSES IN GENERAL. 


Or, as there is no load between C and E, and if np — x 2 is 
the length of load from D towards G , Eq. (7) will become: 



— r 


. •. r = 


l 


P 



1 



Eqs. (6) and (8) will enable the position of moving load to 
be at once computed without trial. 

Example II. —Loads at Panel Points Only. 

If loads are located at the panel points only, then W lf W 2y 
W z , etc., will be the panel loads, and a , b , c , etc., the panel 
lengths, and equal to each other in case those lengths are uni¬ 
form ; the parenthesis in Eq. (2) multiplied by (m + i) will 
also disappear. Substituting R from Eq. (1) in Eq. (2) with 
the last parenthesis dropped, there will result: 

Sh = W t (a -f b + c + • • • + x) 4- 1^ + W 2 (b + c + • • • + x) 

4- 1^ + • • • etc. + IV S (c + • • • + x) ^ + • • • + \V n 
— (JEi + fEj + • • • etc.) (/ 4- 1 ).(9). 


The last term of the second member represents the loads 
between G and D. 

The position of loading for a maximum of S' will, in the 
general case, be determined by trial, by ascertaining at what 
position the second group of positive quantities in the second 
member ceases to increase more rapidly (as the load pro¬ 
gresses) than the negative difference between the first posi¬ 
tive group and the negative last member. This can only 
happen if the panel weights toward, or in the vicinity of 
W n are very heavy relatively to those toward, or in the vicinity 
of W x . If the heavy panel loads are W x and those near it, 



GREATEST WEB STRESSES. 


92 a 

(. e.j if the heaviest panel loads are at the head of the train, the 
following analysis shows the positions for maxima stresses, 
in which, it is to be observed, W l is the rear panel load for 
counter stresses. 

Since {a + b + c 4 - • • • + x) ,(i + l) <(/+/) l, and, hence : 

(a -f b + c + • • • + x) (q + 1 

it is clear that for maxima main stresses the loads must extend 
from the farther end of the span to the main member in 
question. 

Since the counter shear is negative, i. e., opposed in sign to 
the main shear, the negative portion of the second member 
of Eq. (9) must be as large as possible for the maximum 
counter shear, and the positive portion as small as possible. 

1 xi 

Hence the portion W 3 (e 4- • • • + x)^ -f - • • + W n — must be 

omitted, and the load W t placed at the panel point nearest 
the end G ; i. e., the load must cover that portion of the span 
between the counter and the nearest end of the span, for the 
maxima counter stresses. 

Hence, for main web stresses under the assumed con¬ 
ditions : 

.S — | W\ {a + b + c + ••• + -r’) + IV 2 (b 4- c + • • • - 4 - x) 

-{-•••+ lV n x | ^ ... (10). 

And for counter web stresses: 

S — j | W± (a+ b-\-c-\- t -'-\-x)-\- W 2 (b + c + • • • + x) 4 - • • • etc. j 
Q+ 1 )~W+ + • • • etc.) ) 



/ 


• • (”)• 



92 b NONCONTINUOUS TRUSSES IN GENERAL. 


The conditions on which Eqs. (io) and (u) are based are 
precisely the same as if the chords are parallel. In the latter 


. . i sec a I 

case a = oo , h — oo cos a, - ri = ——, 7 

hi l h 


, l + 1 

o, and — ? — = sec a. 
h 


Case II.—The Intersection of Chord Sections Above the Inclined Web 

Member. 

This case is illustrated by Fig. 2, in which let the stress S 
in the member DC be under consideration. The moving load 



is supposed to pass on the bridge from E towards G for the 
main web stress. H is the point of intersection of the chord 
sections, while GH and GD are the distances i and m respect¬ 
ively. All other notation remains precisely as in the previous 
case. The reaction, R , under G , is given by Eq. (1). Bear¬ 
ing in mind that the distance DH is now (i — m), and taking 
moments about H , there will result: 


Sh = Ri — W 1 (i—l + a + b + ••• + x) — W 2 (i —l + b + c+ ••• +x) 
- etc. -{\vNy- + w /~^~ C + etc.) (i-m) . . (12). 

Precisely the same operation which follows Eq. (2) shows 
that the desired condition for a maximum is given by the 
following equation : 


+ IE + If3 + ... + W n — -(IVi + JJ 2 ~h etc.) 

1 

+ {iv,+ iv i+ etc./i—'} . . . (13). 
























GREATEST WEB STRESSES. 


The different portions of Eq. (13) evidently represent ex¬ 
actly the same loads as the same portions of Eq. (4). It is 
also clear, from the same considerations, that if the loads 
extend from E to W lr Eq. (13) gives the position for a maxi¬ 
mum main stress, and a maximum counter stress if they reach 
from G to IV 4 (i. e ., to the weight farthest towards E). 

Eq. (13), like Eq. (4), applies to any system of loads what¬ 
ever, and can be applied to special cases in the same manner. 


EXAMPLE III.— Uniform Load. 


By using the same notation and the same process of reason¬ 
ing as in Ex. 1, Eq. (13) takes the form for the greatest main 
stress: 


n 




Or, for the greatest counter stress: 




EXAMPLE IV.— I,oads at Panel Points Only. 

Let it be first supposed that i is less than /; i. e., i < /. 
The same general considerations that were given in Ex. II., 
applied to Eq. (12), will cause it to take the form : 



— W% {b -f* c 4* • • • "H x} 


Sk = — W t (a + b + c 4 -- 



( 


+ ( IV 1 + W 2 + etc.) (/ — t) . . . (16). 

Since (a + b + c + • • • + x) (/ — i) < 1 ( 1 —i); hence: 


(a + b + c + • 


+ x) (i -£) <(/-*)• 




92 d NONCONTINUOUS TRUSSES IN' GENERAL. 


Therefore, in order that the second member of Eq. (16) 
shall have its greatest positive value, the loads must be at all 
the panel points. Eq. (16) also shows that in the case now 
under consideration there can be no reversal of stress in any 
web member. In the inclined member, S, the stress will 
always be tension, and always compression in the vertical 
member passing through its upper extremity. 

The positions for the actual maxima stresses can be found 
by trial only, as they will depend on the amounts of the panel 
loads and their locations relatively to each other. 

Let it next be supposed that i is greater than /; i. e., i > l. 

In this case (/ — i) becomes negative, or (i — /) positive, 
and the conditions for maxima values are the same as those 
fixed for Eq. (9); hence they need no further attention. 

A remaining example with the intersection of chord sec¬ 
tions belozv the inclined web member, and between it and the 
end of the span, can be treated in precisely the same general 
manner as the preceding. The moving load between G and 
D would lie, in the general case, partially on one side of the 
point of intersection and partially on the other. This form 
of truss, however, has little or no technical interest, and needs 
no further attention. 

The preceding treatment applies to any forms of trusses, 
whether deck or through, with one or either chord horizontal. 
In the application of any particular formula it is only neces¬ 
sary that the point of intersection of the chord sections shall 
be located according to the conditions on which the formula 
is based. 

The treatment also applies to any system of web members, 
whether they are all inclined at different angles to a vertical 
line, or at equal angles; or, again, if, as in the Figures, a part 
of them are vertical. It is only to be borne in mind that two 
web members intersecting in the unloaded chord take their 
greatest stresses together only when that chord does not change 
its direction at that point of intersection. In case that direc¬ 
tion does change, the value of “ i ” will be different for the 
two members, although “ m ” will remain the same. 

The tabulations mentioned in Art. 7 and given in Arts. 9 


GREATEST WEB STRESSES . 


92 ^ 


and 11 can be used in the application of the preceding formulae 
precisely as with parallel chords. Short methods of compu¬ 
tation, well known to every engineering office, make their 
practical applications easy and rapid. 

In designing trusses with variable depth, special care must 
be taken in determining counter web stresses. It frequently 
happens, in cases similar to Fig. 1, that counters must be car¬ 
ried at least one panel nearer the end of the span than paral¬ 
lel chords would require. Again, the vertical web members 
are frequently subjected to heavy tension, with special con¬ 
ditions of moving load, at panel points where the chords 
change direction. 


Numerical Example . 

The truss to be taken in this example is exactly similar to 
that shown in Fig. 1, except the upper chord will be straight 
from the hip to the top of the inclined end posts. 

Let / = 297 ft. ; p = 27 ft. .*. ~ = 11. 

P 

“ m — 3/ = 81 ft. 

“ the centre depth of truss be constant for three 
panels and equal to 42 feet. 

“ the depth of truss at the first panel point from 
the end be 24 feet. 

As the web member whose stress, S, is now desired is in 
the fourth panel, the distance, i, is 81 feet; 1. e. y 3 p. The 
moving load to be used is that shown in Fig. I of Art. 
77, and tabulated on page 41. If this load passes on the 
bridge from right to left, and if the first driving wheel rests 
at the panel point at the right extremity of the panel in 
question— i. e the fourth panel from the left end—93.8 feet 
of the uniform load of 3,000 pounds per lineal foot will rest 
on the bridge in addition to the two locomotives. In Eq. (4) 
the parenthesis ( W\ 4- IV^ + etc.) will, with the above position 
of loading, be equal to zero, while (W 8 + W 4 + etc.) will equal 


92/ NON- CON TIN UO US TRUSSES IN GENERAL. 

either 15,000 or 39,000, or some value between. With the 
values taken in this example, Eq. (4) will become: 

W 1 + W 2 + • • • + W n = 22 (IV 3 + Wt + • • • + etc.); 


and with the position of loading assumed, this equation will 
take the following numerical form : 

623,400 > (22 x 15,000 = 330,000), 

and < (22 x 39,000 = 858,000). 

Hence the position of loading assumed satisfies Eq. (4). 
The resulting reaction at the left end (G, Fig. 1) of the span 
is easily and quickly found by Eq. (1), with the aid of the 
tabulation on page 41, to be: ^ = 215,000 pounds. The 

lever arm, h, of the stress desired is found from the truss 

dimensions already given, and is 151.2 feet. The distance b' 
in Eq. (2) is 18.917 feet. The preceding data substituted in 
Eq. (2), gives: 

Sh — 16,687,458 S = 110,370 lbs. 

The elements of this example have designedly been so 
chosen as to show how greatly the computations may be 
shortened and simplified by a proper choice of the depth, 
chord inclination, and intersection distance, i. 

Art. 21.—Position of Moving Load for Greatest Chord Stresses. 

Since the chord stresses under any given system of loading 
depend only on the truss depths at the various panel points, 
measured in a direction normal to the opposite panels, the 
positions of loading for their maxima values with inclined 
chords will be identical with those determined in Art. 7 for 
parallel chords. No new conditions, therefore, are to be 
developed here. The equations of Art. 7 are to be applied 
exactly as they stand. It is only to be remembered that the 
lever arm for any panel is the normal erected on it to the 
opposite panel point. 


HORIZONTAL COMPONENT OF GREATEST STRESS. 93 


Art. 22.—Horizontal Component of Greatest Stress in any Web Member 
—Constant Value of the Same for Vertical and Diagonal Bracing 
with Parabolic Chord—All Loads at Panel Points. 

Of more interest, perhaps, than real value to the engineer, 
is the expression for the horizontal component of the greatest 
stress in any web member, though it may very easily be 
written. It may be useful at times as a numerical check. 



Let the figure represent a truss of one system of triangula¬ 
tion, subjected to the action of vertical loads passing along 
the lower chord AB . It is desired to find the horizontal 
component of the greatest tensile stress in GH . Let Hh and 
Gg be verticals passing through H and G. 

The following notation will be used : 

p — panel length (uniform) in AB. 

n — number of any panel from B ; for PQ, n has the value 
, 2, and 4 for OH. 

rp — Mg. 
d = Gg. 
d’ = Hh. 

N — number of panels in AB. 

I = AB = Np. 
w — moving panel load. 

R — reaction at B. 
n x p — BM. 


For the greatest tension in GH the moving load must 
extend from B to H. 





94 


NON- CON TIN UO US TRUSSES IN GENERAL. 


The distance of the centre of such a load from B is 


2 


Hence, 




i) 


(«i 


- 0 / 
2 / 


Now let moments be taken about 6. 
Hence, 


(MH) = ( 'r («, 


r)/ - Oi - 




Or, (j«y) = % ['-(«.-*•)/] ( _ I} 

2 d l 



In order to obtain the horizontal component of the stress 
in 67 % due to the assumed load, it is only necessary to take 
moments about H in precisely the same manner. The ex¬ 
pression, however, can be derived immediately from Eq (i) 
by putting r — I, and writing d ' for d. 

.*. Hor. Com. (GF) = ^ —- —-L n x (n x — i) . (2). 

t 

The horizontal component of the greatest tensile stress in 
GH is the difference between the second members of Eqs. (1) 
and (2); let it be called H x . 


... H t =Sg-{ j „ (Wi _ ,) . (3) . 

If a is the angle of inclination of GH to a horizontal line, 
then: 

( GH) — H x sec a .(4). 


Eqs. (3) and (4) apply to all tensile web stresses. For com¬ 
pressive web stresses as typified by ( GM) there would be 
found the Hor. Comp. (GK), instead of Hor. Comp. (GF), 
by taking moments about M; d' would then represent Mm. 
By making r — o in Eq. 1 : 













95 


HORIZONTAL COMPONENT OF GREA TEST STRESS. 



Hence, for the horizontal component of the greatest com¬ 
pressive stress in GM : 



And, (GM) = sec a .... (7). 


By means of the Eqs. (3), (4), (6), and (7), every web stress 
in the truss may be determined by formula. 

If GM is vertical, r = o and d = d' in Eq. (6), and H{ == o, 
as it should. 

If GH is vertical, Eq. (3) shows H x to be zero in the 
same manner. 

It is to be borne in mind, in the application of these formu¬ 
lae, that n is counted along the loaded segment; also that 
d r , for tension, is taken at the head of the train, and one 
panel in front of it for compression. 

If the moving load passes along the upper chord, exactly 
the same formulae hold true, but d' taken at the head of the 
train will give compression, and tension when taken a panel 
length in front. 

If the curve KFC is a parabola, with vertex at the centre 
of the span, if K and C coincide with A and B, respectively, 
and if GM and all corresponding web members are vertical, 
Eq. (3) becomes: 


r _ wp f l — (n t — i)p l — n x p 
1 “ 2/ ( d 1 d 


d 


| («i 


1) . (8). 


From the ordinary equation to the parabola: 


j/ 2 = ax 

1 2 

- = ad,; 

4 


and 







NON- CON TIN UO US TRUSSES IN GENERAL. 


9 6 


in which d x is the depth of the truss at the midle of the span. 
Hence, 


, / 2 

f — —r x - 

4 di 


In this equation put y =- n^p and x = d^ — d, then y — 

- — (n x — i) p and x — d x — d', successively. There will re- 
suit: 

1 - n x Ip + nop 2 = — T (d x — d) ; 

4 4"i 

- - {n i - i) lp + (»i - i) 2 / = 7/W - 
4 4"i 

Remembering that / = : 

v _ j 4 («i jV — «, 2 ) . 

“ — “1-]y 3 -> 

v 4((«1 - i)^V-(»i - i) 2 ) 

a — a x ^y2 

Putting these values in Eq. (8), also Np = /, there will re* 
suit: 


rj _ wp 2 N 2 

8^/ 


iV- (%- i) 
i) (»i -i ) 2 


iV— 
iV — /q 2 



wl 


= constant 


| «! («! - I) 

* » ( 9 )* 


As this is the horizontal component of the greatest tension 
in any diagonal web member and constant, that greatest stress 
itself is the hypothenuse, parallel to the brace in question, of a 


right-angled triangle of which the base is H x — 


wl 










BOWSTRING TRUSS. 97 

This furnishes a very short method of finding the stress in 
any inclined web member. 

The similarity between H x and the total stress in the hori¬ 
zontal chord, with the truss wholly loaded, is interesting. 

If the trussing is so designed that the diagonal or inclined 
braces sustain compression, Eq. (6) gives precisely the same 
general result, but with the sign changed. 

In such a case there would be substituted in the parabolic 

equation y = ^ — n^p and x — d x — d' also, y = — — (n t — 1 )p 

and x — d x — d\ d and d' having changed places. 

If no web members are vertical, y will have for one value 

in the equation to the parabola, — — (n t — r)p instead of 

— — n t p, the other values to be substituted remaining the 
2 

same. This new value gives, 

r , 4 [(«i — r)N - [n x - rf] 
a — "1 * 


Now making the substitutions in Eq. (3) instead of Eq. (8): 



Art. 23.—Bowstring Truss—Diagonal Bracing—Example. 

The first form of bowstring truss to be treated is that shown 
in Fig. 1. All braces are inclined, and each apex in the upper 
chord is vertically over the centre of the panel below. 

The truss is supposed to be designed for a highway bridge. 
There is a sidewalk on either side. 









98 NON-CONTINUOUS TRUSSES IN GENERAL. 

The greatest moving load will be assumed to be that of an 
advancing crowd of people, from the left end of the span, 
weighing eighty-five pounds per square foot. 

As the span is a short one, and the roadway heavy, the 
whole of the fixed load will be put upon the lower chord. 

The following are the data required: 

Span = 72 feet. Depth of truss at centre = 11.7 feet. 

Radius of circumference of circle passing through apices in 
upper chord = 60 feet. 

Number of panels = 6. Panel length = 12 feet. 

Width of roadway, from centre to centre of trusses = 20 feet. 

Width of each sidewalk = 6 feet. 

W — 900 pounds per foot = 5-4 tons per panel. 

w — 32 x 85 x 12 — 16.32 tons per panel. 


As usual, W and w refer to fixed and moving loads respect¬ 
ively. 



In all the diagrams that follow, the lines indicated by any 
two letters are parallel to the members of the truss at the 
extremities of which the same letters are found. 

In this truss and in the two which follow, the upper and 
lower chord sections found in any panel intersect outside of 
the span RR r , hence the positions of the moving load, for the 
greatest web stresses, are precisely the same as those which 
would be taken for a truss with parallel chords. 

With the head of the moving load at M, the truss is first 
to be considered as divided through the members AB, BM , 
and ML ; then through BC, BL , and ML. 




BOWSTRING TRUSS. 


99 


Fig. 2 is the complete diagram for this position. 

R (reaction) = 27.1 tons. 

{ML) = {R x 18 — 21.72 x 6) -f- 9.3 = 38.4 tons. 9.3 feet is 
the depth of truss through B . 



The shear is, 

5 = 27.1 — 21.72 = 5.38 tons. 

The diagram needs no explanation. It gives: 

{BM) — + 18.5 tons. {BL) — — 2.1 tons. 

With the head of the moving load at L: R = 37.98 tons. 
Fig. 3 is the complete diagram. 



The truss is first supposed to be divided through BC , CL, 
and LK; then, through CD , CK , and LK. 

{.LK ) = (R x 30 - 2 x 21.72 x 12) -r- 11.7 = 52.8 tons. 
The shear is, 

5 = 37.98 - 43.44 = - 546 tons. 








IOO NON CO it TIN UO US TRUSSES IN GENERAL. 

The diagram gives: 

( LC ) = + 20.0 tons. ( CK ) = — 0.60 tons. 

With the head of the moving load at K: R = 46.14 tons. 
Hence, 

(KH) = (R x 42 — 3 x 21.72 x 18) -7- 11.7 = 65.4 tons. 
Fig. 4 is the complete diagram. 



The shear is, 

5 = 46.14 — 65.16 = — 19.02 tons. 

The diagram gives: 

(KD ) = + 21.7 tons. {DH) = — 7.4 tons. 

With the head of the moving load at H: R' = 40.7 tons; 
and S = — (40.7 - 5.4) = - 35.3 tons. 

{HG) =(R' x 18 - 5.4 x 6) -r 9.3 = 75.3 tons. 


ct 



Fig. 5 is the diagram, and it gives: 
(HE) = + 20.9 tons. 

; 

( t 


(EG) — — 4.0 tons. 







BOWSTRING TRUSS. 


IOI 


With the head of the moving load at G, or with the moving 
load over the whole bridge : R = 2.5 x ( W + w) = 54-3 tons. 
In this case no chord stress is found by moments, but the 
diagram, Fig. 6, is worked up from the end of the truss. 


a (Rid:) 



It gives: 

( RA) = — 100.1 tons. 
( AB ) = — 109.1 “ 

(BC) = — 103.0 “ 
(CD) = — 102.0 “ 


(RM) = 4- 84.6 tons. 
(ML) = + 93.6 “ 

(. LK) — + 97.0 a 
(AM) — + 19.5 “ 


The results of these diagrams are collected and written in 
Fig. 1. 

Both web and chord stresses may be checked by moments 
as follows. 

Moments about K give : 

(CD) = - (54.3 x 36—2 x 21.72 x 18) -f- 11.7= — 100.2 tons. 

The diagram gave 102.0 tons. The agreement is close 
enough for the purpose, but in an actual truss the difference 
ought not to be greater than one per cent, of the smallest 
result. 

Again, BC and LK intersect in a point about 30.8 feet to 
the left of R, and the normal distance from that point to CL 







102 


N0N-C0NTINU0US TRUSSES IN GENERAL . 


produced is about 48.5 feet. Hence, with the head of the 
moving load at L, and, by taking moments about the point of 
intersection: 

(CL) = (2 x 21.72 x 48.8 — R x 30.8) -f- 48.5 = + 19.6 tons. 

The diagram gave + 20.0 tons. The agreement is suf¬ 
ficiently close. 

Numbers of checks like the two above should be applied. 

The Fig. 1 shows that the web members MB, BL , CL , CK, 
DK, DH , HE, and EG must be counterbraced. 


Art. 24.—Bowstring Truss—Vertical and Diagonal Bracing with 

Counters. 

• 

The case next to be taken is that of an ordinary “ bow¬ 
string” truss with vertical and diagonal bracing, and is rep¬ 
resented in Fig. 1. The inclined braces are not supposed 
capable of resistance to a compressive stress. Inasmuch as 
counters are almost invariably introduced in such a truss in 
ordinary engineering practice, they will be supposed to exist 
in this case. 



The truss is supposed to be designed for a highway bridge, 
furnished with sidewalk on each side. 

The greatest moving load will be taken as a crowd of peo¬ 
ple weighing eighty-five pounds per square foot, covering 
roadway and sidewalks, advancing panel by panel from R 
until the truss is entirely covered. 

Since the span is a short one, and the roadway very heavy, 
the whole of the fixed load will be taken as applied to the 
lower chord. 









BOWSTRING TRUSS. 


103 

The following are the data required ; they are taken from 
the preceding Article: 

Span = 72 feet. Depth of truss, 12 feet. 

Radius of circumference of circle passing through upper ex¬ 
tremities of verticals = 60 feet. 

Number of panels = 6. Panel length = 12 feet. 

Width of roadway, from centre to centre of trusses = 20 feet. 

Width of each sidewalk = 6 feet. 

W = 900 pounds per foot = 5.4 tons per panel. 

w — 32 x 85 x 12 = 32,640 pounds = 16.32 tons per panel. 

As in Art. 18, if the plane dividing the truss cut more than 
three members, some one of these members must be neg¬ 
lected or assumed not to exist. 

For the sake of brevity, two letters inclosed by a paren¬ 
thesis will denote the stress in the member indicated by these 
letters. 

The placing the load for the greatest web stresses, is done 
in accordance with the general principles established in Art. 
20. 

When the head of the moving load is at Z, the existence 
of AK must be ig¬ 
nored ; for if BL be 
then omitted, AK will 
suffer compression. 

With the head of 

the train at Z, R= 27.1 _ 

' Fig. 2. 

tons; hence 

(LK) — (R x 24 — 21.72 x 12) -4- (BK — 10.8) = 36.1 tons. 

Fig. 2 is the complete diagram for this case, and explains 
itself, ac is the shear 2 P = 27.1 — 21.72 = 5.38 tons. Scaling 
from Fig. 2 : 

« 

(BL)= + 19.2 tons. (BK) = + 2.0 tons. 

BK is also omitted for this loading. 





104 


NON- CON TINU0 US TRUSSES IN GENERAL. 


With the head of the moving load at K, R = 37 - 9 ^ t° ns i 
and CG and BH are omitted. ( KH ) = (R x 36—2 x 21.72 x 


c 



Fig. 3. 

18) -f- ( CH — 12) = 48.78 tons. Fig. 3 is the diagram for this 
case, ce is the shear, acting downwards, 2 P= 37.98 — 2 x 
21.72 = — 5.46 ton. 

This diagram gives the results : 

( CH ) = — 1.0 ton. {KC) — + 16.0 tons. 

With the head of the moving load at H , = 46.14 tons. 

DF and CG will be omitted. 

It is unnecessary to give the diagram for this case. It is 
drawn precisely as the two preceding ones have been. 

The diagram gives: 

( DG ) = —1.0 ton. (HD) — + I 7 A tons. 

Neither will the diagram for the head of the load at G be 
given, as it is constructed exactly like the others. It gives, 
omitting GC and DF: 

(GE) = + 15.00 tons. 

For the greatest chord stresses the moving load covers the 
entire truss, and Fig. 4 of the next Article is the complete 
diagram for the case. All the explanation which the diagram 
needs is there given. BL, KC, GC and FD are supposed to 
be omitted. 

Taking moments about B , with uniform load (w + IV): 

(KH) — (54.3 x 24 — 21.72 x 12) 10.8 = 96.5 tons. 

Others may be checked in the same way. 

The most rational circumstances under which the greatest 
tensile stresses can be supposed to occur in the verticals, are 





BO WSTRING TRUSS. 


105 


those under which they are found in the next Article ; and 
the results there obtained are used in this case. They are 
introduced, without more explanation, in Fig. 1. Their dia¬ 
grams will be found in the next Article. 

These results are by no means satisfactory, but nothing 
better can be done with such a form of truss. 

Some of the web stresses should be checked by moments, 
by the general method. 

Fig. 1 shows the greatest web stresses selected from all the 
preceding results. 

It is far more convenient to treat the fixed and moving 
loads together, as has been done in this case, than to treat 
them separately, as, of course, may be done. 

Again, the stresses caused by each panel load on all the 
members of the truss may be found, and their effects com¬ 
bined, but this also requires far more labor than the method 
followed. 

The stress diagram for each position of the moving load 
might have been worked up from the end of the truss, as was 
that for the chord stresses, but it saves considerable labor to 
find one chord stress by moments, and begin the diagram 
with that. 


Art. 25. — Bowstring Truss — Vertical and Diagonal Bracing without 

Counters. 


It is evident, from what has preceded, that the existence of 
the counters causes considerable ambiguity in the web stresses, 



and it is much more satisfactory from a strictly technical 
point of view to leave them out, as shown in Fig. 1. 







106 N0N-C0NTINU0US TRUSSES IN GENERAL. 

.Fig. i is exactly the same as Fig. i of the preceding Arti¬ 
cle, with the counters omitted, and in this Article will be 
found the stresses existing in it with precisely the same data 
as were used above. 

The moving load is brought on panel by panel from R , ac¬ 
cording to the general principles established in a preceding 
Article. 

With the head of the moving load at L, R = 27.1 tons. As 
before, (. LK ) = 46.9 tons. 

CL 

b 

Fig. 2. 

Fig. 2 is the complete diagram for this loading. (RL) = 
(LK) — 46.9 tons, and ab — 27.1 — 21.72 = 5.38 tons. 

Hence, (AK) — 11.9 tons. 

With the head of the moving load at K, R = 37.98 tons. 

(KH) — (R x 24 — 12 x 21.72) -7- 10.8 = 60.27 tons. 

Fig. 3 is the diagram for this loading, and it explains itself. 
Hence, 

(BK) = + 24.9 tons. (BH) = — 15.8 tons. 


Fig. 3. 

It is unnecessary to show the diagrams for the two cases of 
the head of the moving load at H and G . They give respect¬ 
ively : 









BOWSTRING TRUSS. 


IO7 


(DH) — + 17.1 tons, ( DG)= — 1.0 ton, and ( GE ) = 4- 15.0 tons. 

Fig. 4 is the diagram for the moving load over the whole 
bridge, bd is the reaction R = R' = 54.30 tons ; dc — 21.72 
tons ; and fg is equal to J (21.72) tons. 

The greatest chord stresses, taken from Fig. 4, are written 
in Fig. 1. 

The greatest web stresses are selected from all the preced¬ 
ing results, and also written in Fig. 1. 

The stress ( BC ) may easily be checked by moments, as 
follows, by taking the origin at H. The normal distance of 
H from ( BC ) is 11.9 feet. Hence, 

(BC) = {2^ (w + W) x 36 — 3x12 (w + W) \ -T-11.9=98.56 tons. 

Other and similar checks should also be applied. 

It is seen in Fig. 1 that the diagonals need counterbracing, 
but there is no ambiguity, and the superiority of the design 
over that in the preceding Article is evident. 



It is to be carefully borne in mind that the diagrams must 
always be drawn as large and as accurately as possible. Those 
of the present Article were constructed to a scale of ten tons 
to the inch. The figures do not show the scale. 






io8 


NON-CONTINUOUS TRUSSES IN GENERAL. 


Art. 26.—Deck Truss with Curved Lower Chord, Concave Downward 

—Loads at Panel Points—Example. 

The truss shown in Fig. i is, in some respects, a peculiar 
one. It has one prominent characteristic which distinguishes 
it from the bowstring trusses which have been treated in the 
three preceding articles, in that the chord sections (upper and 
lower) in any panel, excepting those two at the centre, inter¬ 
sect in the upper chord within the limits of the span. All 
the web stresses, except those in DN and DM, will have their 
greatest values when the moving load covers the whole truss, 
as was shown in Art. 20. NM is horizontal, consequently the 



intersections of NM with CD and DE are found at an infinite 
distance from the truss, but ON and CD intersect between E 
and A (near the latter point): all other intersections are found 
between A and G. 

The positions of moving load for the greatest stresses in 
DN and DM are the same, therefore, as for a truss with 
parallel chords. 

Observations relating to the positions of the points of inter¬ 
section of the chord sections in the figure apply to a truss for 
which the following are the data: 

Radius of circumference of circle passing through the apices 
of the lower chord = 60 feet. 

Vertical distance of centre of circle below D — 66 feet. 

Depth of truss through D — 6.3 feet. 

Span = AG = 72 feet. 

AR — GR! = 18 “ 






DECK TRUSS . 


109 


Uniform upper chord panel length = 12 feet. 
NM — 12 feet. 

ON = ML =13 “ 

OP =LK = 11.7 “ 

Uniform panel fixed load = 5.4 tons = W. 
“ “ moving load = 16.32 “ = w. 



The loading is the same as that used in the preceding bow¬ 
string trusses. 

Let the angle which DN or DM makes with a vertical line 
be denoted by a. Then, 

tan a = 0.952 ; sec a — 1.38. 

For greatest tensile stress in DN. 

With the moving load extending from A to C: 

Reaction R = 37.98 tons. 

The shear .S' = 2 x 21.72 — 37.98 = 5.46 tons. 

Hence, (DN) — S sec a = +7.53 tons. 

For greatest compressive stress in DM. 

With the moving load extending from A to D: 

% 

Reaction R = 46.14 tons. 

The shear 5 = 46.14 — 3 x 21.72 = — 19.02 tons. 
Hence, (DM) = S sec a — — 26.25 tons. 







IIO 


NON- CON TIN UO US TRUSSES IN GENERAL. 


For the other web stresses the moving load must cover the 
whole truss, and Fig. 2 is the complete diagram for that con¬ 
dition of loading. With the data given below the figure, no 
explanation is needed. The results of the diagram will be 
found in Fig. i, together with those determined above by the 
trigonometrical method. 

One advantage inherent in this form of truss, as in all in 
which the chord intersections are found within the limits of 
the span, is the little counterbracing required. 

In the truss taken, the two web members DN and DM are 
all that require such treatment. Indeed, with a sufficiently 
small radius of lower chord and centre depth, together with 
an odd number of upper chord panels, a truss may readily be 
designed which will require no counterbracing at all. A dis¬ 
advantage, however, is the small depth at centre, just where 
a great one is needed, with the resulting heavy chord stresses. 

As the Fig. i shows, no stress exists in PR or KR' ; never¬ 
theless those members would ordinarily be inserted for the 
purpose of stiffening the whole structure. 

The truss may be supported directly, as at G , or there may 
be an end post, as AR. 

The greatest stress in AR, will be the reaction R added to 
J (w + W). Hence, 

(AR) = — (54.3 + 10.86) = — 65.16 tons. 

As checks, moments about D give: 

(NM) — (54.3 x 36 - 2 x 2172 x 18) 6.3 = + 186.0 tons. 

The normal distance from C to ON is 7.5 feet (nearly). 
Hence, by moments about C : 

(ON) — (54.3 x 24 — 21.72 x 12) 4 - 7.5 = + 139.0 tons. 

OP, prolonged, cuts the upper chord at a point about three 
feet from D toward E, and the normal distance from that 
point of intersection to OB, prolonged, is about 23.5 feet. 
Hence, 

(OB) — (54.3 x 39 - 21.72 x 27) -f- 23.5 = 4- 65.1 tons. 


CRANE TRUSSES. 


111 


The agreement of the last result with that obtained by 
diagram is very close, but the other results, by the two 
methods, show a difference of about two per cent. This is 
close enough for the present purpose, but in practice the 
figure and diagrams should be drawn large enough to make 
this difference, at most, one per cent, of the smallest result. 

A truss of this character, with more than one system of 
triangulation, gives indeterminate stresses, but the approxi¬ 
mate method of Art. 18 may be used. Approximate deter¬ 
minations may also be made by treating each system, with its 
weights, by the methods just given, and combining the results 
for the chords. 


Art. 27.—Crane Trusses. 

A form of truss which has been used for powerful cranes, 



Fig. i. 








112 


NON-CONTINUOUS TRUSSES IN GENERAL. 


under circumstances requiring much head room, is that shown 



in Fig. i. It revolves about a vertical axis midway between 
E and F. 





CRANE TRUSSES. 


113 

In the example taken, the weight, W, hanging from D, the 
peak, is supposed to be ten tons. 

Each chord of the truss m , /, k, etc., or a , b, c y etc., is made 
up of chords of quadrants of two circumferences of circles. 
The radius for the chord mlk, etc., is 25 feet, and that for 
the other chord is 22.6 feet. EF is 5 feet. 

Denoting the chord panels by single letters: 

a = 5 feet. 
m = b = c = 6 “ 
l — d— 7 “ 
k=e=S “ 
h — 9 “ 

Fig. 2 is the complete diagram for the stresses in the truss, 
supposing the only load to be the ten tons hanging from the 
peak. If it should be necessary to take into account the 
weight of the truss, it would be done precisely as the fixed 
weights of trusses have been treated in the preceding Articles. 

The lines in Fig. 2, denoted by letters and figures, are par¬ 
allel to lines denoted by the same letters and figures in Fig. 1. 

The diagram gives the following results: 


(I) = 

+ 

12.5 

tons. 

( 2) = 

: — 

14 

tons. 

(3) = 

+ 

6.00 

<< 

(4) = 

: — 

i 3-7 

a 

(5) = 

+ 

1.50 

u 

( 6) = 

: — 

134 

a 

( 7 ) = 

— 

2.60 

it 

( 8) = 

: — 

14.0 

u 

( 9 ) = 

— 

7.20 

it 

(10) = 

: — 

13.2 

u 




(II) = 

— 5.8 tons. 




(*) = 

— 

8.7 tons. 

(m) = 

= + 

17.8 

tons. 

(*) = 

— 

24.0 ' 

u 

U) = 

= + 

304 

u 


— 

35-9 • 

u 

{k) = 

= + 

38.7 

a 

(d)= 

— 

44.0 

u 

(*) = 

= + 

44.2 

a 

(e) = 

— 

48.2 1 

u 

(g) = 

z 4 * 

47.2 

a 




(/) = 

— 54.1 tons. 





These results may easily be checked by moments. The 


1 14 NON-CONTINUOUS TRUSSES IN GENERAL. 

different lever arms, with two exceptions, to be used, are 
shown in Fig. i. The normal distance from g to C is about 
4.6, and from e to B about 5.3 feet. These lever arms were 
scaled from the drawings, and may not be exactly right, but 
near enough for the purpose. By moments about C : 

( g ) = + (10 x 22.25) -r 4-6 = 4 - 48.4 tons. 

By moments about B : 

( e) — — (10 x 26.0) -4- 53 = — 49.0 tons. 

The chord sections d and k, prolonged, meet at A , and mo¬ 
ments about that point give : 

(7) = - (10 x 3.7) -4- 14.7 = - 2.5 tons. 

These results agree sufficiently well with those obtained by 
the diagram. 

If the chain, rope, or cable pass along either chord, the ten¬ 
sion in it will tend to produce an equal amount of compres¬ 
sion in the panels of that chord. The resultant stress, there¬ 
fore, in any panel will be the algebraic sum of this amount of 
compression, and the stress due to the weight W. 

Fig. 3 is a skeleton diagram of the ordinary crane, which 
revolves about the centre line of ED as an axis. AB is the 

weight hung at the peak, A. BC 
is parallel to DA , and represents 
the amount of tension in that 
member. AC is the compression 
in AE due to the weight W. As 
before, the tension in the rope or 
cable tends to produce an equal 
amount of compression in any 
member along which it lies. 

Let / denote the normal dis¬ 
tance from DA to any point in 
the centre line of DE ; then any 
section of DE will be subjected to the bending moment: 

M= (DA) x /. 


XL 







ROOF TRUSSES . 


115 

DE will also be subjected to a direct stress (tension in 
Fig- 3) equal to the vertical component of the stress in DA. 
The greatest resultant intensity of stress in any section will 
be the combination of the intensities due to these two causes. 


Art. 28.—Preliminary to the Treatment of Roof Trusses—Wind Pressure 

—Notation. 

Four, only, of the principal types of roof trusses with 
straight rafters will be treated, since the method used for 
any one is precisely the same in character as that to be used 
for any other. 

The wind will be assumed to act on one side of the roof, 
and its resultant action will be assumed to be normal in direc¬ 
tion to the rafters. If such is not the case, f and v will 
represent empirical determinations of the horizontal and 
vertical components, respectively, in the Articles which fol¬ 
low, and the methods will remain precisely the same. 

Let p be the intensity of this normal wind pressure, and let 
/ and 4 be the lengths of two adjacent panels of the rafter, 
while d is the horizontal distance between two adjacent and 
parallel rafters. The total normal wind pressure supposed 
exerted or concentrated at the point between the two panels, 

will then be + A) . or if / — 4, as is usually the case, pdl. 

If 6 is the angle which the rafter makes with a horizontal 
line, the horizontal component of this normal pressure will be: 

f — & ^ {J- +J_i ) s - n q . or s i n Q . 

and the vertical component will be: 

p d (l 4 - 4 ) 


v — 


cos 6 ; or pdl cos 6 . 


Finally, l&t the total fixed weight of the roof be supposed 
concentrated at the panel points of the rafters; and let the 
weight of such load at any panel point be represented by W 2 . 
The total vertical load at any panel point will then be: 

W= W 2 + v. 

The vertical reactions due to the vertical component of the 





NON-CONTINUOUS TRUSSES IN GENERAL. 


116 


wind pressure and the fixed load are found by the principle 
of the lever in the usual manner ; that at the left of the span 
will be called R, and that at the right, R'. 

The vertical reactions due to the horizontal forces f, will, 
however, be called R ", and they will be found for the different 
cases by taking moments about any point in the horizontal 
line joining the feet of the rafters. If b is the vertical pro¬ 
jection of a rafter, and 2 c the span, or distance between the 
feet of two rafters meeting at the ridge, there will result: 


R" = bp 



b 2 p 

4 C ' 


At the foot of the rafter pressed by the wind, this reaction 
will be downward in direction ; at the foot of the other rafter 
it will be upward. 

The total horizontal reaction at the points of support will 
be equal to bp, the total horizontal force of the wind, and its 
direction of action will be opposite to that of the wind. The 
horizontal component of the wind pressure and the horizontal 
reaction produce a couple, equal and opposite to that whose 
force is R", and whose lever arm is 2 c. 

R" must be numerically less than R, or the wind will turn 
over the roof bodily. 

If the foot of one rafter is supported on rollers, the hori¬ 
zontal reaction will be wholly exerted at the foot of the other. 

If the foot of neither rafter is supported on rollers, the 
horizontal reaction will be assumed to be equally divided 
between the points of support. 

The stresses for the vertical and horizontal loads are found 
by separate diagrams, although they might be found by one 
only, because the slope of the roof may, in some cases, be so 
small as to make it needless to consider the forces f 

If rollers are used at the foot of one rafter, the wind may 
press that one or the other. In treating a large roof it may, 
then, be necessary to take the wind first in one direction and 
then in the other. 

These two, with the case of no rollers, make three possible 
cases, and an example will be taken in each one. 



ROOF TRUSSES. 


II 7 


Art. 29.—First Example. 

The truss represented in Fig. i is a roof truss, applicable 
to short spans. There is no “ moving load ” in such a case. 
The wind pressure, however, may act on one side and not on 
the other, and for that reason W, W u and W 2 are taken as 
differing from each other, as was explained in the preceding 
Article. 



There is no essential error in this case in assuming all the 
load concentrated at the points indicated. The wind press¬ 
ure is assumed to act on the left side of the truss, so that 
W > W t > W 2 . Also, W x = } (W + W 9 ). A and B are equal 
in length, and E is horizontal. 

Figs. 2 and 3 are the stress diagrams for Fig. 1, and the lines 
in them, indicated by letters primed, are parallel to, and repre¬ 
sent stresses in the members marked with the same letters in 



Fig. I. The kinds of stresses in the different members are 
shown by the signs + or —, in both figures, signifying tension 
or compression, respectively. 











I 


11 8 NOiV- CON TIN UO US TRUSSES IN GENERAL. 

Fig. 2 is the diagram for the vertical loads, and Fig. 3 that 
for the horizontal loads f. 



Rollers are supposed to be under the foot of neither rafter. 
Consequently the horizontal reaction at each end is as 
shown. f_ is equal to \ f. 

In Fig. 1, R' acts downward at the left of the span, and 
upward at the right. 

The resultant stress in any member is the algebraic sum 
of those given by the two diagrams. 

Fig. 3 does not show all its lines parallel to the members of 
Fig. 1. There is, of course, a diagram similar to Fig. 2, for 
the right half of the truss, but it is not needed. 

The stresses may also be found by the method of moments, 
by locating the origin of moments according to the general 
principle stated in Art. 17. 


Art. 30.—Second Example. 

Fig. 1 of this Article represents a very common roof truss. 
As before, the total load is supposed concentrated on the 
rafters at the points indicated. Wind pressure is taken as 
acting on the left of the roof, making W > W l > W 2 . W 2 is 
simply the panel weight of the roof, and W 1 = £'(W+ W 2 ). 
Each rafter is divided into four equal parts, and W is taken 
equal to 2 W 2 . Hence the reaction R = 3 W. This does not 
at all affect the generality of the diagram. The lower ex- 





ROOF TRUSSES . 


11 9 

tremity, however, of D\ in Fig. 2, will not usually be found 
at a. 



Fig. 2 is the stress diagram for the vertical loading taken, 
and the notation has precisely the same meaning as before. 

It is seen from Fig. 1, that there is some ambiguity in 
regard to the stresses C', Q', F' f and M’. It may be assumed, 
however, that E' = H' and P' — Q\ which makes F' — 
2 H' =2 E\ 



The kinds of stresses are shown on the diagrams. 

















120 


N0N-C0NTINU0US TRUSSES IN GENERAL . 


Fig. 3 is the diagram for the horizontal stresses f and 

Rollers are supposed to be placed at the foot of A. Hence 
all horizontal reaction will be found at the foot of Ay. As 
shown, that reaction will be equal to 4 f. The vertical reac¬ 



tion R" will be directed downward at the foot of A, and 
upward at the foot of Ay. 

Considering the horizontal forces only: 

a / = b; = Ci = Di = — A'. 

Ky = Ly — a = - 4 / — K\ 


The resultant stress in any member is found by combining 
the results of the two diagrams in the usual manner. 

This style of truss is so frequently used that formulae for 
the stresses due to the vertical loading are given below, in 
which a is the length of the rafter, c the half span RR\ and b 
the height of Wy above O. These expressions may be readily 
derived from Fig. 1. 

A ' = J R > B' = A' — W b —, 


C' = A' -2 W 


a 


D' = A r 



y 



F' = 2 H' = 2- W. 

a 








ROOF TRUSSES. 


121 


P' ~ Q' = hH'j = hjWy 
L' = K'— P' = j (R — \ W), 
O’ = U - M'= j (R - f W), 


M'=XF' — = — W 
2 b b ’ 

N' = M' + < 2 ' = 4 t 


Art. 31.—Third Example. 

Figs. I, 2, and 3 are roof-truss and stress diagrams, respect¬ 
ively. The wind pressure is supposed to act on RW it and 
the total load is taken to be concentrated as shown. 




The rafters are each divided into three equal parts. 









122 


NON-CON TIN UO US TRUSSES IN GENERAL . 


The notation has the same signification as that used be¬ 
fore, and it is unnecessary to explain the diagrams. 

Fig. 2 is the diagram for the vertical loading, and Fig. 3 
that for the horizontal forces f and f = if 

Rollers are supposed to be placed at the foot of A x in Fig. 
1. Hence the total horizontal reaction will exist at the foot 
of A f and its value will be 3 f as shown. As before, the 
direction of R" will be upward at the foot of A lf and down¬ 
ward at the foot of A. 

The resultant stress in any member will be found by com¬ 
bining the results given in Figs. 2 and 3. 



Fig. 3. 


Fig. 3 does not show K' equal to Hf as it is not a scale 
diagram. 

The members FH and K are in tension, while D and E are 
in compression. W> W[> W 2 , the last being the fixed panel 
weight of the roof. 


Art. 32.—Fourth Example. 


In this Article, as before, Fig. 1 is the truss, and Fig. 2 is 
its stress diagram for the vertical loading. Wind pressure is 

w; 



Fig. 1. 










GENERAL CONSIDER A TIONS. 


123 


taken as acting on RW 1 ; the reaction R , therefore, is i (3 F 7 + 
fFi + IF 2 ) and JV > > fF 2 . The rafters are each divided 

into three equal parts, and F and D are vertical. is the 
fixed panel weight of the roof, and W x — (W + W 2 ) 2. 



The kinds of stresses in the various members are shown in 
Fig. 2. 

It is unnecessary to give the diagram for the horizontal 
components of the wind pressure. The method of drawing 
it will fall under some one of the three cases already given. 


Art. 33.—General Considerations. 


In all the preceding cases of roof trusses the stresses may 
be obtained by the method of moments, and therefore the 
graphical results may be checked by that method. 

The operations are precisely similar if a part of the load is 
hung from points in the tie RR ', in all the cases, or if the 
loading is even more eccentric than that assumed. 

In many cases the sides of large buildings are braced to 






124 


NON-CONTINUOUS TRUSSES IN GENERAL. 


the roof by oblique members extending from* that panel point 
of RR' adjacent to the side, to some point in that side. In 
such cases the wind will cause stresses, in the different mem¬ 
bers of the roof truss, which must be determined independ¬ 
ently of those already found and added, algebraically, to 
them. 


CHAPTER IV. 


SWING BRIDGES. ENDS SIMPLY RESTING ON SUPPORTS. 

Art. 34.—General Considerations. 

WITHOUT regarding the nature of the supports or attach¬ 
ments at the extremities of the two arms, swing bridges are 
divided into two classes: those with center-bearing turn¬ 
tables, and those with rim-bearing turn-tables. In the first 
class the entire reaction at the pivot pier is exerted through 
a central pivot, or a nest of one or more series of solid, con¬ 
ical rollers; usually the latter. In such a case there may be 
a circular drum or framework, supported on wheels running 
on a circular track, but they are used solely for the purpose 
of steadying the bridge while open. 

In the case of a rim-bearing turn-table, however, the reac¬ 
tion at the pivot pier is exerted through the circular track on 
which the wheels supporting the drum or framework of the 
table turn. The object of a pivot, in such a case, is simply 
to enable the bridge to turn truly about a center. 

In reference to the truss, there is evidently only one point 
of support, at the pivot pier, with a center-bearing turn-table, 
i. e. y at the center. 

With a rim-bearing turn-table, however, there may be two 
or more points of support at the pivot pier, though it will be 
shown hereafter that, by separating the different systems of 
triangulation, it will never be necessary to consider more than 
two at once. 

Again, with either turn-table there may be three different 
methods of supporting or securing the extremities of the two 
arms of the bridge. These extremities may simply rest on 
supports, so that the reaction will always be zero, or upward; 

125 


126 


SWING BRIDGES. 


this is the only case which will receive more than a passing 
notice in this chapter. 

Again, those extremities may be fastened down or latched 
to the piers when the bridge is not open. The reaction may 
then be nothing, upward or downward. 

In these two cases the reactions at the extremities of the 
arms will be zero when the bridge is simply closed, and sup¬ 
porting no moving load. 

In the first, when the moving load is on one arm, the ex¬ 
tremity of the other may be slightly raised from its support; 
in the second case, however, that extremity will be held down 
by the latching apparatus, i. e., the reaction will be downward. 
The object of the latching apparatus is thus seen to be the 
prevention of the hammering of a truss-end on its support. 

Finally, the third method is to raise, by proper machinery, 
the truss ends, when the bridge is closed, any desired amount. 

The object of this arrangement is to insure a reaction at 
the extremities, which will always be nothing or upward, and 
thus obviate the hammering before mentioned. 

In this case the whole of the bridge weight does not rest at 
the pivot pier, as the lifting of the ends takes up a part of it; 
in fact, may take up the whole of it. 

Recapitulating, then, the ends of a swing bridge may be: 

(i.) Simply supported, 

(2.) Latched down, 

(3.) Lifted up. 

The detailed consideration of the first case will next be 
taken up. 


Art. 35.—General Formula for the Case of Ends Simply Supported—Two 
Points of Support at Pivot Pier—One Point of Support at Pivot 
Pier. 

With two points of support at the pivot pier there usually 
arises the case of a continuous beam resting on four points of 


>4 l 7 

A - 


Til5.1 
33 l 2 C 






GENERAL FORMULAE. 


127 


support, as shown in Fig. 1 . The notations of the spans and 
bending moments at the different points of support are suf¬ 
ficiently well shown in the figure. The points of support 
will all be taken in the same horizontal line, as the formulae 
will then also apply to any configuration belonging to a state 
of no stress, provided the truss may be considered straight 
between any two points of support (see Appendix). Any 
truss may be considered straight when an equivalent solid 
beam has a neutral surface which is plane before flexure ; 
a straight solid beam is “ equivalent ” to a straight truss when 
equal moments of inertia and resistance are found at the same 
section in the two structures. 

The theorem of three moments, in the ordinary form, does 
not apply, then, to a continuous truss with one chord curved, 
and none of the following investigations apply to such a 
case. 

Again, in the span L there will be supposed no load, as 
such is usually the case. The load on 4 ought always to be 
supported on short girders or beams resting at B and C> for 
there is the less complication of stresses in the trusses, and 
consequently less liability to uncertainties ; besides, such an 
arrangement is probably more economical in material. 

In the present case and M± will each be equal to 
zero. 

Let z denote the distance of the point of application of any 
force, P, from the left-hand end of the left-hand span, or 
right-hand end of right-hand span. In l ly z would be meas¬ 
ured from A, while in 4 it would be measured from D. 

The formula expressing the theorem of three moments for 
all supporting points in the same level becomes, by using the 
notation of Fig. 1 : 

M& + 2 Mlk + li) + M % k + 7 W - #)g + - *) z = °- 

4 4 

1 2 

The symbols 2 and 2 indicate summations for the spans 
4 and / 2 . 

Applying the above equation to spans 4 and 4, and then 


128 


SWING BRIDGES. 


to 4 and 4 there will result, bearing in mind the circumstances 
of the present case: 

2M 2 (/, + 4) + M, 4 + iXP( 4 2 -z*)z = o . (i). 

4 

Af 2 4 + 2J/3 (4 + 4) + 4 2 — z*)z = o . (2). 

4 


If Equation (1) be multiplied by 4 > and Equation (2) by 
2 (4 + 4) *> and if the results so obtained be subtracted, and 
if the following notation be used: 

h. 




4 ’ 


b= k. 

/ ’ 
L 2 


z 

n =7 ; 
4 


z 

m — — 

4a 


there will result: 




A^Pi^i — n 2 ) n — 2^4 + 1) ^/^(i — w 2 ) w 


4 (^ + 1) (c + 1) — 1 
Equation (1) then gives: 

Af 3 + Al 2 ^P(i — n 2 ) n 




2{C + I) 


( 3 ). 


( 4 )- 


Equation (2) will evidently give another expression for 
M 2 , but it is not necessary to write it. 

Let R lf R 2y R 3 , and R 4 be the reactions at the points of 
support A, B, C, and D respectively, Figure 1. Then, adapt¬ 
ing the formulae for reactions from the theorem of three mo¬ 
ments (see Appendix) to the notation of the present case, 
there may at once be written 

1 M 

R,= 2 P{ 1 -n) . . . 

4 



R . = 2 Pn - 


Mo Mo - Mo 


R3 — + 


4 

M, - M* 


L 


s +2Pm- J ^ . 


4 4 

3 Mo 

R,= 2 P(i -m) +9? . 

4 

As should be the case, there is found: 


( 6 ). 

( 7 > 

( 8 ). 








GENERAL FORMULAE. 


129 


In ordinary swing bridges where 4 = 4 = /, and a single 
weight P rests on 4 : 

Ri-P{ 1 -») j 1 + " )n 


4 if + 1 ) - 


( 5 «> 


c + I 


RI = R(l — « 2 ) n — -- 

4(c + i) 2 - r 

R i = c\R 1 -R i -P{i-n)}-R i 

R 2 — P — R\ — R 4 4- P3 . . 


. . (6a). 

■ ( 7 «)- 

. (8a). 

It may sometimes be convenient to use the following equa¬ 
tion derived from Equation (1): 

M 3 = —{2M 2 (c + 1) + c 2 l l 2 2 P( 1 — ft 2 ) n\ . (9). 

These are all the equations necessary for the solution of 
the case of two supports at the pivot pier, frequently exist¬ 
ing if the turn-table is rim-bearing. 

If there is only one point of support at the pivot pier, the 

Ti £2 

3 ^ Zc 


A 


--* 

case reduces to that of a continuous beam of two spans only, 
as shown in Figure 2 . 

As A and C are points of support only, M x and M s are each 
zero; hence if /ft now stands for the equation immediately 

4 

preceding Equation (1) gives: 


— 4 


R'2P( 1 — ft 2 ) ft 4- 2 P(i — /ft 2 ) /ft 


[2 ~ 2 2 (c 4- 1) 

There will also result: 

Ri = 2P(l — n) + ~ . . 

4 

R 2 =2Pn - ^ +SPm - ^ 

2 M 

R a =2P{i-m) + J -p . 

4 


• • 


(10) . 

(11) . 

(12) . 

(13) - 







SWING BRIDGES ! 


130 


If 4 = 4 = /, then £ = 1 and Equation (10) becomes: 

M 2 = — | — ?r) « + ^/’(i — m 2 ) w | . (13#). 

There is again found: 

R x + R 2 + R z = 2 P + XP. 

1. 3 

These complete the general formulae needed for the case of 
ends supported. 

Some very important deductions are to be drawn from 
Eqs. (5), (6), (7), (8), (11), (12), and (13), considering them 
applied to bridges with rim-bearing turn-tables. 

Those equations are so written that a positive value of R 
means a reaction upward in direction, while a negative value 
indicates a downward reaction. 

In the case of Fig. 1, let the span l s be supposed free of 

3 

any loads P, then the term involving the summation 2 will 
disappear. Eq. (3) then shows that M s will always be posi¬ 
tive ; consequently Eq. (4) shows that M % will always be 
negative. 

Using these results in connection with Eqs. (5), (6), (7), and 
(8), it is at once seen that R 2 and R 4 will always be positive, 
while R s will always be negative. 

It may also be shown that Eq. (5) makes R x positive in such 
cases as always arise in an engineer’s practice, although that 
equation apparently shows that R x may, under some circum¬ 
stances, be negative, since M 2 is always negative in the case 
taken, while (4 — z) is, of course, always positive. 

By deducing the value of M 2 from Eqs. (3) and (4), and 
introducing it in Eq. (5), there will result: 


*i = 


4L 


2 P (4 - z) - 


2 (4 + 4) 


4 h (4 + 4 ) (4 + 4 ) — 4 ' 


2P (4 s - z*) 


If 4 is very small in comparison with 4 or l s , and if, at the 





GENERAL FORMULAE. 


131 


same time, z is small, there may be written for a single “ P,” 
nearly: 





which is evidently positive. 

If, on the contrary, / 2 is small and z large, there may be 
written for a single weight P, nearly: 




P(h + *)(h 



In this expression R x can be equal to zero only by suppos¬ 
ing the negative quantity within the brackets to be larger, 
numerically, than it ought to be, i.e., by making (4 + z) = 24, 
and z — 4 ; hence it can never be negative. 

If 4 = 4 —/ s = l the general value of R x takes the form : 



P{l-z) 


15/ 


X P{P-*)Z 


If z is small, there results nearly: 




If z is large, nearly : 




p(/-z)-Fp(i 



In neither case, therefore, can the reaction be negative. 

It may, consequently, be assumed as a principle that if / 2 is 
small in reference to 4 or 4 or if it is equal to those quan¬ 
tities, the reaction R h in the case supposed, must always be 
positive; and within those limits must be found all cases of 
swing bridges. 

Since any load on the span 4 makes the reactions at A and 
D positive, considered by itself, so any load on / 3 will, of 
















132 


SWING BRIDGES. 


itself, make the reactions at D and A positive. Consequently, 
as there is never any load on L, the reactions at A and D will 
always be positive, and the ends of the bridge will never tend to 
rise from their points of support. No “hammering,” there¬ 
fore, can take place, in this case, at the ends. 

The case of the two points of support, B and C, taken in 
connection with a center-bearing turn-table will be considered 
further on. 

Fig. 2 represents the case of either a center or rim-bear¬ 
ing turn-table with only one point of support at the pivot 
pier; the two cases are coincident in all their circumstances. 

Eq. (io) shows M 2 to be always negative. Consequently, if 
there is no load on / 2 , R x and R 2 will always be positive, while 
R 3 will always be negative. 

When span 4 carries load, however, the span 4 may, at 
the same time, support just enough load to make R 3 equal 
to zero; more load than that will make R 3 positive, or upward 
in direction. 

But in the present case the point C is simply a point of 
support , consequently no negative or downward reaction can 
exist there. It becomes necessary, therefore, to determine 
just how much load on / 2 , combined with the full load on l lt 
will make R 3 = o. For this purpose, must be taken 
from Eq. (io) and inserted in Eq. (13), while in the latter R 3 
must be just equal to zero. For the sake of brevity, let 

j 2 P ( 4 2 — P) z be represented by A. 

4 

Then from Eq. (10): 



A _1 

2 ( 4 d~ 4 ) 24 (4 + 4 ) 


2P ( 4 * 


z*) z . (14). 


A is a constant quantity so far as this operation is con¬ 
cerned. 

Putting this value of M 2 in Eq. (13) after making R 3 — o: 


24(4 + 4 ) 2 P (4 - z) - 2 P ( 4 * - z 2 )z = Al 2 . (15). 
Eq. (15) indicates what disposition of the load on the span 




ENDS SIMPLY SUPPORTED . 


133 

/ 2 will make R 3 = o ; and it will be seen to be of very easy 
application. 

The determinations of R lt R 2 , and R 3y for all loading in 
excess of that indicated by Eq. (15) will require the use of 
Eqs. (11), (12), and (13) as they stand; for all loading less 
than that amount, however, R 3 = o, and the reactions R l and 
R 2 are to be found by the simple principle of the lever, con¬ 
sidering 4 as a simple overhanging arm, or cantilever. These 
operations will be shown in detail hereafter. 

In this case it is evident that “hammering ” will take place 
at the ends with certain dispositions of loading. 

Art. 36.—Ends Simply Supported—Two Points of Support at Center— 

Partial Continuity—Example. 

The general formulae of the preceding Article will be 
applied to the truss shown by skeleton diagram in Figure 1 of 
Plate X, in which the intermediate verticals P 2 and P 3 , with CP 
and the inclined web members EP and P x , are in compression, 
while the remaining vertical and diagonal web members are 
in tension. 

As shown by the diagram, the total length of the trusses 
between the centers of end pins is 295.5 feet. The following 
statement gives all the dimensional data required in the suc¬ 
ceeding computations: 

Center depth at CP — 42 feet. 

Depth “ T x = 32 “ 

“ “ P 2 = 3 of “ 

“ “ E s = 29L “ 

“ “ T 10 = 28 “ 

Length of center panel = 20 feet 6 inches. 

Panel length = 27.5 feet. Total length = 295.5 feet. 

Point of intersection of upper and lower chords of arms is 
20 panels from the free extremity of each arm. 

The trusses are placed 16 feet apart, centers transversely. 

The trigonometric quantities required are: 


134 


SWING BRIDGES. 


tan 

for EP 

= 0.982 

sec 

for 

EP = 

1.4 

u 

a 

a 

= 0.048,5 

a 

a 

a = 

1.001,21 

<< 

a 

P 

= 0.364 

a 

a 

P = 

I.064 

« 

a 

T t 

= 0.938 

a 

u 

T t = 

i- 37 i 

a 

a 

T,> 

= 0.897 

a 

<< 

T s = 

1-343 

a 

il 

T, 

= 0.859 

a 

u 

E = 

I- 3 I 9 

a 

ti 

cc 

= 0.976,2 

u 

u 

cc = 

1-397 


The moving load to be used consists of two coupled con¬ 
solidation locomotives, identical with that shown in Figure I 
of Article 77, followed by a uniform train load of 3,000 pounds, 
or 1.5 tons, per lineal foot, the moment tabulation for which 
is shown on page 41. 

The fixed load will be as follows : 


Track . 

400 pounds per lineal foot. 

Floor beams and stringers . 

320 “ 

<< U il 

Trusses. 

N* 

V* 

O 

CO 

On 

<< <( tt 

Total.. . 

1,700 “ 

ti it a 


The weight of trusses will be taken to be divided equally 
between the upper and lower panel points, but the track, 
floor beams, and stringers will be taken wholly at the latter. 
Hence the upper and lower panel fixed loads will be as fol¬ 
lows : 

1 « 

Lower panel load ~~~ x 27.5 = 16,650 lbs. = 8.325 tons. 

Upper “ “ x 27.5 = 6,750 “ = 3.375 “ 

Total “ “ 23,400 “ = 11.700 “ 

In order that the weight of the locking gear at the extrem¬ 

ity of each arm may be properly provided for, the weight con¬ 
centrated at the free extremity of each truss will be taken at 
5 tons, which is somewhat more than one-half of each of the 
other lower panel fixed loads. 

The panel load at the top of CP will be: 


27.5 4- 20.5 490 


x -— 5,880 lbs. — 2.94 tons. 


2 


2 














ENDS SIMPLY SUPPORTED. 


135 


The stresses due to the fixed loads given above will first be 
found. Since it is assumed that the ends of the arms are sim¬ 
ply supported when the draw is closed, the stresses arising 
from the fixed or own weights are cantilever stresses identi¬ 
cal with those of the open draw ; or, in other words, the entire 
fixed load is carried at the two points of support over the 
drum on the pivot pier. As the chords of the arms are not 
parallel, all web stresses will be found by taking moments 
about the point of intersection of those chords, distant 20 
panels from the extremity of either arm, as stated above. 
For example, in order to find the stress in T 3 , an imaginary 
cut or section is to- be taken through b, T 3 , and 3 ; it is then 
to be noted that the lever arm of T 3 {i.e., the normal dropped 
from the intersection of b and 3 on T 3 produced) is 22 x 27.5 
-r- sec for T 3 . Now the lever arm of the five-ton weight at the 
extremity of the arm is 20 panels ; that of the panel loads act¬ 
ing along T w , 21 panels ; and that of the loads acting along 
P 3 , 22 panels. 

Hence moments give: 

(T s ) — - -•----- sec T 3 = 36.815 tons. 

The chord stresses will be found by the same method of 
moments, the center of moments being at the panel point 
opposite each chord panel— i.e., for (b) the moment center is 
at the intersection of/^and T 3 ; and at the intersection of T z 
andi^ for (3)- These methods are precisely those outlined in 
Article 17, excepting that of the diagram, which is not used 
here. 

It is to be observed that all the stresses can be found by 
the graphical methods given in the preceding pages, but the 
moment method only will be used in this example. 

It will be assumed that the counter c x sustains no fixed load 
stress. 

Under the preceding explanations, the desired stresses 
will be as follows: 


(1) = — 5 tan for EP = — 4.91 tons. 



136 


SWING BRIDGES . 


, . 5 x 55 + 11 .7 x 27.5 

(2^ = — --—---— = — 20.35 tons. 

V ; 29.33 


( 3 ) =- 


5 X 82.5 + II .7 (55 + 27 - 5 ) _ 


44.93 tons. 


( 4 ) = (s) = - 


30.67 

5x110+11.7 (82.5 + 55 + 27.5) 


32 


= — 77.52 tons. 


By taking moments about top (or foot) of CP'. 

(6) _ _ _ _ 5 X 137.5 + n -7 (no + 82.5 + 55 + 27.5) 

' ' e 42 

= — 92.98 tons. 

(a) = — (1) sec a — 4 - 4.92 tons. 

( b ) = — (2) sec ot — + 20.37 “ 

M = - (3) seca= + 44.98 “ 

( d ) = e sec = + 98.93 “ 

(EP) — 5 sec for EP = + 7 tons. 

m = - (Si» + iSZSfi) = - a ,4 ton. 


«) = - 2 


5 X 20 + II.7 x 21 + 3 375 x 22 


22 


= — 19.09 tons. 


. _ 5 X 20 + 11.7 (21 + 22) + 3,375 X 23 

' 2 23 

= — 29.595 tons. 

Since chord d cuts the lower chord 3.2 panel lengths from 
foot of 7 i, or 0.8 panel length from the end of the arm, the 
center of moments for P x will be at that point, while the lever 
arm for the same member will be (4.2 panel lengths ~ sec P x ). 
Hence: 

/ox 11.7 (0.2 + 1.2 4 - 2.2 4 - 3.2) — 5 x 0.8 

4.2 

= - 17*99 x l - 3 l 9 = — 23.73 tons. 












ENDS SIMPLY SUPPORTED. 


137 


The stress in cP is simply the vertical component of ( d ) 
added to the fixed load at the top of cP. Hence: 

(cP) = — (e) tan /3 — 2.94 = — 92.98 x 0.364 — 2.94 

— — 36.785 tons. 

The stresses in the tension members T 4 , T Zi and I ; can 
now readily be found by the diagram, Figure 2 of Article 17, 
as could those also in the vertical compression web members ; 
but the moment method will be continued, and in using it the 
center of moments must be taken at the point of intersection 
of the chords, twenty panel lengths to the left of the foot of 
EP. The lever arm of T 4 will be (21 panel lengths ~ sec T 4 ), 
and similarly for the other inclined tension members. Hence: 


. 5x20+11.7x21 _ 

(A) = + - — - sec 7 * = 4- 22.57 tons. 

X 



+ 


5 X 20 + I 1.7 (21 4- 22) 
22 


sec T 3 = + 36.815 tons. 


(T,) = + 


5 X 20 + 11.7(21 4- 22 4- 23) 


sec = 4- 50.02 tons. 


( 7 j) = fixed load at its foot — 4- 8.325 tons. 


Since there is no fixed load stress in the counter^, and as the 
members LS' and cc are designed simply to steady the bridge 
when open, and, hence, can sustain no stress that can be 
computed, the preceding computations complete all the fixed 
load truss stresses. 

The moving load stresses are next to be found, and in 
finding them the condition of “ partial continuity” will be 
assumed— i.e., it will be assumed that no shear can ever pass 
from one arm of the bridge across the center panel to the 
other. The web members in the center panel are light in 
section, and proportioned only to hold the bridge in a steady 
condition, when open, against any tendency to vertical vibra¬ 
tion by the wind, or irregular or uneven operation of the turn¬ 
ing machinery. These vibration stresses are entirely indeter- 





138 


SWING BRIDGES. 


minate and cannot be computed, but it is known, from actual 
experience with drawbridges of great length, that they are 
small, and amply resisted by rods and braces quite inadequate 
to carry more than a very small portion of the moving load 
shear that would, under some conditions of loading, pass the 
center panel if the trusses were perfectly continuous over the 
drum. Hence such a construction is designed as will not 
permit the existence of those moving load shears at the cen¬ 
ter, thus leaving only the incidental stresses of turning to be 
cared for. 

The effect of this arrangement is assumed to make either 
arm of the bridge a simple truss supported at each end for 
all moving loads on that arm, so long as the other arm carries 
no moving load. 

It should be observed that the span / 2 (ordinarily termed 
the center panel), included between the central points of sup¬ 
port, sustains no load. 

If the moving load partially covers both arms, or the whole 
of one arm and a part of the other, it is to be divided into 
three parts. Two of these parts, together with the reactions 
R x and due to them, produce equal and opposite moments 
at the center. For these two parts, consequently, the truss is 
one of perfect continuity, with the main diagonals at the 
center omitted. This last condition is admissible, because for 
these two parts the shear at the center will be zero. 

For convenience these two parts will be called “balanced” 
while the third part will be called “ unbalanced.” 

For the unbalanced part, the arm or span in which it is 
found will be a simple truss supported at each end. 

If the panel loads are uniform in amount, balanced loads 
will be symmetrically placed in reference to the center. 

If the panel loads are not uniform in amount, the balanced 
portions would be determined by equating M 2 to M % , with the 
aid of Equations (3) and (4) of Article 35. 

Coupling these statements with the principles deduced in 
the preceding articles, it will at once be seen that the greatest 
reaction R x at A, Figure I, Plate X, will exist with the moving 
load over the whole of the left arm, for any moving load on 


ENDS SIMPLY SUPPORTED. 139 

the right arm will balance a part of that on the left, and relieve, 
to some extent, the reaction at A. 

Although the moving load consists of the various wheel 
concentrations followed by the uniform train, as shown by the 
following diagram ; and while these concentrations can read¬ 
ily be used for the moving load on one arm only, according 
to the principles of Articles 20 and 2i‘, their use in the same 
general manner for balanced moving loads would lead to 
excessive complication. Hence, for the latter loads, a fixed 
series of panel weights, resulting from one position of the dia¬ 
gram concentrations given above, will be used in the corre¬ 
sponding stage of the computations. By this device the 
determination of the stresses will be much simplified, and the 
results will be essentially accurate. 



From what has preceded, it is clear that the greatest coun¬ 
ter stresses are to be found by carrying the moving load for¬ 
ward from the center of the bridge toward the end, and on 
one arm only, at the same time considering it a simple non- 
continuous span. 

The greatest main web stresses, on the other hand, are to 
be determined by moving the system of balanced panel 
weights, already described, from the two extremities of the 
bridge toward the center, as will presently be illustrated. 

It is to be remembered that verticals in compression will 
sustain their greatest stresses at the same time with the 
diagonal tension members which cut their upper extremities, 
if the moving load traverses the lower chord. 

As one arm of the bridge is a simple truss supported at 








140 


SWING BRIDGES. 


each end, for the unbalanced loads on it, it is evident that the 
greatest compression in the upper chord and tension in the lower 
will exist, near the ends, for the moving load over the whole of 
one arm. 

Since moving loads on both arms at the same time balance 
each other, it results that the greatest tension in the upper chord 
and compression in the lower, at the center, will exist with the 
moving load over the whole of both arms. 

This is true for the center only. For other panels adjacent 
to the center, it will be necessary to take single-balanced panel 
moving weights, and find for each, all panels in which the stress 
is of the same kind as that caused by the fixed load alone , and 
the amount of that stress in those panels. 

Having obtained the results for each pair of balanced 
weights, they are to be combined in the manner already 
shown. 

The greatest compression in the lower chord and tension 
in the upper, near the ends, however, will exist with the bridge 
open or closed and subjected to its own weight only. 

From these results the greatest chord stresses are to be 
found. 

The maximum moving load compression in EP will next 
be found, and then the counter stresses and that in T 10 , for 
all of which the arm of the draw will be treated as a non-con- 
tinuous span, with the moving load passing over it from the 
center toward the end. The positions of the moving load for 
the greatest value of these stresses will, hence, be found by the 
principles of Articles 7, 20, and 21. Equation (7) of Article 
7 shows that the third driver of the front locomotive must be 
placed at the foot of Z* 10 in order to give the greatest shear in 
EP, or, what is the same thing, the greatest reaction at its 
foot. Such a position of the moving load will make the shear 
in question 45.16 tons. Hence : 

(EP) = — 45.16 x 1.4 = — 63.23 tons. 

In order to determine whether the counter c x is necessary, 
let it be supposed omitted, and the resulting greatest moving 


ENDS SIMPLY SUPPORTED. 


141 

load compression found in T x . Equation (4) of Article 20 
shows that the latter will exist with the third driver of the 
front locomotive at the foot of P 3 , whence will result a reac¬ 
tion at the foot of EP and a load at the foot of T 10 of 32.1 
tons and 5.72 tons, respectively. By taking moments about 
the point of intersection of the chords, and using the same 
lever arm for T 4 as was employed for the fixed load stresses: 



32.1 x 20 — 5.72 x 21 
21 


sec T 4 = — 24.85 x 1.371 


= — 34.08 tons. 


But the fixed load tension in has already been found to 
be 22.57 tons, which is much smaller than the 34.08 tons of 
moving load compression just found. Hence, as T\ can really 
sustain no compression, the counter c x must be introduced, and 
its greatest tension found with the same position of moving 
load, and, also, under the assumption that 7 " 4 does not exist. 
Remembering that the lever arm of c x is 22 panel lengths 
divided by the secant for c x : 



32.1 x 20 — 5.72 x 21 
22 


sec c x = + 23.72 x 


1.4 


= 4- 33.21 tons. 


By a precisely similar use of Equation (4) of Article 20, it 
will be found that the greatest moving load compression will 
exist in T s with the first driver of the front locomotive at the 
foot of P 3 . This position will give a reaction of 13.05 tons at 
the foot of EP, and an advance load of 11.02 tons at the foot 
of P 3 . Again taking moments about the point of intersec¬ 
tion of the chords, the greatest moving load compression in 
r 8 will be found to be : 



13.05 x 20 — 11.02 x 22 
22 


sec T 3 = — 14.45 tons - 


But the fixed load tension in T 3 has already been found to 
be 36.815 tons, which is over two and a half times the 14.45 
tons of moving load compression. Hence no compression 





142 


SWING BRIDGES. 


can ever exist in 7\, and no counter will be needed in the 
same panel with it, nor, indeed, any other counter than c x . 

The same condition of loading will give the greatest mov¬ 
ing load tension in P 2 , but computations will show that it is 
only about one-third the fixed load compression (as would 
be inferred from the results for T s ); hence no counterbrac¬ 
ing of that member is required. 

These operations show the method always to be pursued 
in determining the requisite counters, and, also, that it is 
essentially identical with that used for non-continuous spans. 

As T l0 is a simple hanger, it will take its greatest tension 
with the greatest floor beam reaction at its foot. This latter 
will exist with the same position of moving load as that 
which gives the greatest stress in EP — i.e., with the third 
driver of the front locomotive at the foot of and its value 
is 25.57 tons. Now the shear existing at the same time in 
EP has been found to be 40.16 tons, and as the co-existing 
vertical component of the stress in a is between three and four 
tons only, the counter c x must act in tension to carry sufficient 
shear over to EP to make up its shear of 40.16 tons. In 
other words, the vertical components of (c x ) and ( a ) added to 
the stress in T 10 must equal 40.16 tons. Hence the stress in 

Tio is simply the greatest possible load at its foot. 

-*• 

.-. (r i0 ) = + 25.57 tons. 

The member d sustains no stress, under the condition of 
partial continuity, with the moving load on one arm only; 
hence it is to be ignored in finding the stress in P x . The 
greatest compression in this latter member will therefore 
exist with precisely the same position of moving load as was 
used for EP, but changed end for end on the span, so as to 
be headed toward the center rather than toward the end of 
the arm. 

By using the same shear as for EP, therefore: 

{Pi) — — 45 - t 6 x sec P x — — 59.57 tons. 

The tension member T x is a simple hanger, with the maxi- 


ENDS SIMPLY SUPPORTED . 


M3 


mum floor beam reaction at its foot, as was found for T lQ . 
Hence: 

T\ = + 25.57 tons. 


The moving load will pass on the arm from the end 
toward the center, for the remaining web members. The 
position of moving load for T ’ 2 , in accordance with Equation 
(4) of Article 20, requires the third driver to be at the foot of 
P 2l with both locomotives and tenders, but no train load, on 
the arm. This position gives a reaction of 32.1 tons at the 
foot of P lf and an advance weight of 5.72 tons at the foot of 
T x . Hence: 



32.1 x 25 — 5.72 x 24 
23 


sec T 2 = + 38.16 tons. 


In the same manner it is found that the greatest tension 
in T s will exist with the second driver of the first locomotive 
at the foot of P 3 , and with the corresponding driver of the 
second locomotive on the arm and nine inches from its end. 
The reaction at the foot of P 1 will be 15.24 tons, and the 
weight at the foot of P 2 , 3.14 tons. Hence, by moments, as 
usual, about the point of intersection of the chords : 



15.24 x 25 - 3.14 x 23 
22 


sec T s = 4- 18.85 tons. 


Since P 2 sustains its greatest compression under the same 
condition of moving load as T 3 : 



15-24 x 25 
23 


= — 16.57 tons. 


Again, Equation (4) of Article 20 shows that the first driver 
of the front locomotive must be at the foot of T x0 in order to 
give T 4 its greatest tension. The second wheel of the tender 
will then be on the arm and one foot from its end. Hence 
the reaction at the foot of P t will be 4.62 tons, and the 
weight at the foot of P 3 , 1.1 tons, and : 



4.62 x 25 — 1.1 x 22 


sec Ti = H- 5.96 tons. 


21 






144 


SWING BRIDGES. 


The same position of moving load gives the greatest com¬ 
pression in P %; hence : 



4.62 x 25 
22 


5.25 tons. 


These results complete the moving load web stresses for 
the moving load on one arm only, leaving the chord stresses 
yet to be determined. 

The lower chord panels I and 2 receive their greatest 
stresses with EP, and they are to be found by simply multi¬ 
plying the shear in that member by the tangent for EP. 
Hence: 

(1) = (2) — 45.16 x 0.982 = + 44.345 tons. 


Similarly, the panels 4 and 5 sustain their greatest stresses 
with P x . Hence: 


(4) — (5) = 45.16 x 0.859 = 4 - 38.79 tons. 

In order to determine the greatest stress in lower chord 
panel 3, recourse must be had to Equation (14) of Article 7. 
That equation shows that, with the moving load passing on 
the arm from right to left, the third driver of the second 
locomotive must be at the foot of P 2 , with 24 feet of the 
uniform train load resting on the arm adjacent to the center. 
This position gives a bending moment about the top of 
P 2 (the center of moments for panel 3) of 3,672,530 foot¬ 
pounds, Hence: 



3.672.530 

30.666 


= + 121,070 lbs. = + 60.535 tons. 


The upper chord stresses at once result from those in the 
lower, but it is first to be observed that since T 4 is a tension 
member only, it will not be stressed with the moving load on 
essentially the whole of the arm, since the counter c x will then 
come into action. The upper chord stresses in a and b will 
therefore be equal, and as the foot of P 3 will be the center 




ENDS SIMPLY SUPPORTED. 


145 


of moments for those stresses, the position of the moving 
load for their greatest value will be precisely the same as for 
(3), with the exception that the direction of motion is to be 
reversed, thus placing the third driver of the second locomo¬ 
tive at the foot of P s . Hence : 

, v /rx 3,672,530 

(a) = (b) = - 29.333 seca= “ I2 5,350lbs. = - 62.675 tons. 

Finally: 

(c) = — (3) sec (x ~ — 60.61 tons. 

These computations determine all the greatest moving 
load stresses in one arm considered as a simple non-continuous 
span. 

In finding the greatest stresses due to the moving load on 
both arms, the condition of partial continuity requires the 
use of balanced loads— i.e., loads simultaneously on each arm 
which, if the trusses were continuous over the center, would 
produce equal and opposite center bending moments. The 
most obvious and simple balanced loads are those of equal 
magnitude placed at symmetrical panel points in each arm, 
and such balanced loads will be used in the following com¬ 
putations. 

As has already been observed, the use of the locomotive 
concentrations and the ordinary moment diagram for the 
greatest stresses in this case, would lead to excessive compli¬ 
cation and great labor without any corresponding advantage. 
Essential accuracy can be attained by computing a system 
of panel concentrations or weights from a position of the 
moving load which will give results differing in no sensible 
degree from those determined by the most refined calcula¬ 
tions. The desired position is largely a matter of judgment, 
but a heavy concentration should be found at the panel 
point covered by the head of the train. In the present 
instance the third driving wheel of the front locomotive will 
be placed over a panel point which will be called panel 1. 
A small concentration will, of course, exist at the panel point 



146 


SWING BRIDGES. 


in front of panel 1 ; this will be called the “ advance load.” 
The desired concentrations will then be as follows: 


Advance load, 
At panel 1, 


u 

a 

it 


ti 

tt 

n 


2 , 

3 > 

4 , 


5.70 

25-55 

17-75 

25-55 

19-35 


tons. 

u 

u 

a 

n 


The advance load is sometimes neglected in finding the 
web stresses, as the resulting computations are somewhat 
simplified, and the small error committed is on the side of 
safety. 

All reactions in this case must be found by the equations 
of the preceding Article for perfect continuity, as all loads 
are to be balanced. If a panel load, P, rests on the left-hand 
arm, or span, l u the reaction R t will be by Equation (5 a) of 
Article 35 : 



1 — (1 + n) n 


2 c 


4 0 + 1) - 


C + I 



And by Equation ( 6 a) of Article 35, the reaction R i will 
be: 


R a — P( I — n 2 ) n 


c 

4(y + i) 2 — 1 



In this case, c — 137.5 20.5 = 6.7073. 

If the panel points be numbered from 1 to 4 from the end 
of the arm to the center, panel 1 being at the foot of T l0 , and 
panel 4 at the foot of 7 \; and if P be the general value of a 
panel load, n will have the values 0.2, 0.4, 0.6, and 0.8 in the 
formulae for R x and R±, and the expressions for those reactions 
will be: 


Panel load P at panel I, R x = 0.716102 P; R± = 0.005443 P. 

“ “ “ “ 2, Ri = 0.453178 P ; ^4 — 0.009526 P. 

“ “ “ a 3 , Xi = 0.232204 P; R a = 0.010886 P. 

“ “ “ “ 4 j ^1 = 0.074153 P\ R 4 = 0.008165 P- 





ENDS SIMPLY SUPPORTED. 


H7 


Since balanced loads only are to be employed in this case, 
it will be necessary to determine the reactions R x and R± with 
a panel concentration on one arm balanced by an exactly 
equal concentration symmetrically placed on the other. The 
reactions for each such concentration will be those given 
above, but R x and R± will be interchanged with the exchange 
of one arm for the other. When the two concentrations are 
simultaneously placed on the two arms— i.e., when they are 
balanced—the reaction at each end of the bridge will equal 
that at the other, and will be represented by the sum of R t 
and R±, as given above for a single panel load. If R is that 
reaction : 


For panel I, 
“ “ 2, 


u 

u 


a 

u 


3 > 

4 > 


R = R x + R 4 = 0.7215 P. 
R — R t + R a = 0.4627 P. 
R = R t -Y R i = 0.2431 P. 
R — Rt + R± = 0.0823 P. 


The preceding numerical work can be thoroughly checked 
by finding M 2 for both arms fully loaded, from Equation (4) 
of the preceding Article (having previously found M s ), and 
using it in Equation (5) of the same Article. The result¬ 
ing value of R l = R± should equal the sum of the preceding 
values of R , because both results are for both arms fully 
loaded. The sum of the four values of R is 1.5096 P, while 
Equation (5) of Article 35 gives R { = R = 1.509 7 P. Hence 
the check is satisfactory. In this verification all panel loads 
P are assumed equal, but this does not in any way affect the 
numerical work which it was desired to verify. 

As the reactions for each of the panel loads, and the panel 
loads themselves, are now known, all stresses can readily be 
found. In all the cases except those of simple hangers, or 
where a trigonometric multiplier only is needed, the method 
of moments will be employed, precisely as with the fixed load 
and with the moving load on one arm only. 

The greatest stress in EP will evidently exist with the great¬ 
est reaction at its foot— i.e., with the moving load over the 
whole of both arms. Hence the reaction is : 


148 


SWING BRIDGES. 


0.7215 x 25.55 = 18.435 
0.4627 X 17.75 = 8.210 
0.2431 x 25.55 = 6.210 
0.0823 x 19.35 = 1.595 

.*. R — 34.450 tons. 

Hence: (. EP) — — 34.45 x sec EP — — 48.23 tons. 


The same condition of loading gives the greatest stress in 
lower chord panels 1 and 2 ; therefore : 

(1) = (2) = 34.45 x tan EP — + 33.83 tons. 

The hanger T 10 will be treated precisely as in the case of 
moving load on one arm only, but with the balanced concen¬ 
trations. Hence: 

(T10) = + 25.55 tons. 

The position of loading for the greatest stress in c 1 is with 
the advance weight 5.7 tons at the foot of T 10 , as the resulting 
reaction is then greater in comparison with the load between 
the end of the arm and foot of c l than with any other position. 
If the advance load were very large it would have to h 
placed at the foot of c x for the greatest stress in that me 
ber. The reaction then becomes : 

0.7215 x 5.7 = 4.112 tons. 

0.4627 x 25.55 — 1 1.82 “ 

0.2431 x 17.75 = 4-3*5 “ 

0.0823 x 25.55 = 2.103 “ 

R = 22.350 “ 


Remembering that T 4 is now to be neglected : 



22.35 x 20 — 5.7 x 21 
22 


see c x — 4- 20.83 tons. 


Since the stress in c x is much less than was found w r ith the 
moving load on one arm, as was to be anticipated, it is clear 
that no other counter stresses need be found. Indeed, it is 
evident from the general conditions of the two cases that 





ENDS SIMPLY SUPPORTED. 


149 


the greatest counter web stresses must be found with the 
moving load on one arm only, as has already been observed. 

The conditions of loading for the greatest stresses in the 
main web tension members T 4 , T z , and jT 2 will be found by 
bringing the moving load on the bridge from the end of the arm 
toward the center, every panel concentration at the same time 
being balanced. In the present case, the advance load of 5.7 
tons is to be placed one panel in front of the foot of the mem¬ 
ber in question, as it is small in comparison with that which 
follows it. This advance load might be so large as to require it 
to be placed at the foot of the member whose stress is sought. 
The choice between these two positions can only be deter¬ 
mined by trial. One of these positions of the moving load 
will give the greatest stress desired, for the reason that the 
end reaction will then be the least in comparison with the 
moving load between it and the member considered, thus 
insuring the greatest possible shear in the latter. 

It is to be observed that the design of the truss is such 
that the inclined web members are subject to pure tension, 
and the vertical posts P 9 and P 3 to pure compression, 
necessitating the introduction of the counter c x . If it is 
desired to omit the latter member, making a rather more 
excellent design from a purely engineering point of view, 7 " 4 , 
and perhaps other adjacent inclined web members, will be 
subjected to compression, with possibly some of the vertical 
posts in tension, and all such members must be counterbraced. 
The requisite computations for this case will be considered at 
the end of this Article. 

In order to find the greatest stress in 7 " 4 , the advance load 
of 5.7 tons is to be placed at the foot of P 3 ; hence the 
reaction R will be: 

0.7215 x 25.55 = 18.435 
0.4627 x 5.7 = 2.637 

R z=z 21.072 tons. 



25.55 x 21 — 21.072 x 20 


21 


set T± = +7.51 tons. 





SWING BRIDGES. 


150 

The advance load must be at the foot of P 2 for the greatest 
stress in T 3 , and the resulting reaction will be : 

0.7215 x 17.75 — 12.808 
0.4627 x 25.55 = 11.820 
0.2431 x 5.7 = 1.385 

R = 26.013 tons. 

. 25-55 x 22 + 17.75 x 21 - 26.013 x 20 _ ^ 

. . ( 1 3) ~ ^ 1 3 

= + 25.315 tons. 

For the greatest stress in the advance load must be at 
the foot of 7 i; whence : 

0.7215 x 25.55 - 18.435 
0.4627 x 17.75 — 8.212 
0.2431 x 25.55 = 6.213 
0.0823 x 5.7 = 0.470 

R = 33-330 tons. 


••• (T t ) = 


25.55 X 23 + 17.75 x 22 + 25.55 X 21 - 33.33 X 20 


23 


sec T% — + 48.64 tons. 




In illustration of the effect on the stress in T 2 of placing 
the advance load at its foot, rather than at the foot of T u 
the following would be the value of that stress, rememberin 
that the reaction R would be 26.013 tons, as found for T s : 

, 't'\ t _ 5-7 x 23 + 25.55 x 22 + 17.75 

V 1 V — 


X 21 — 26.013 X 


23 


sec T 2 = + 31. 


which is seen to be about two-thirds of the greatest stress. 
The result would have been very different, however, if the 
5.7 tons were displaced by a large concentration. 







ENDS SIMPLY SUPPORTED. 


151 

As T x is a simple hanger, it receives only the greatest load 
at its foot: 


(~i) = + 25.57 tons. 

The greatest reaction at the foot of P x will exist when the 
advance load of 5.7 tons is placed at that point, and its value 
will be : 

0.7215 x 19.35 = 13.9 62 

0.4627 x 25.55 = 11.820 

0.2431 x 17.75 4.31-5 

0.0825 X 25.55 = 2.103 

R = 32.200 tons. 

Since the upper chord d cuts the lower chord panel I at 
a point 0.8 panel length from the foot of EP, that point of 
intersection must be taken as the center of moments for P x ; 
hence : 


(U = 

32.2 x 0.8+ 19.35 x 0.2 + 25.5 5 x 1.24-17.75 X 2.24-25.55 X 3.2 

4.2 


sec P l = — 56.875 tons. 


The posts P 2 and P 3 take their greatest stresses with the 
web members T 3 and 7 " 4 respectively ; hence the reactions 
for the latter members are to be used for the former. There¬ 
fore : 



25.55 x 224-1775 x 21— 26.013 x 20 

22 


= — 18.025 tons. 


Also: 



25.55 x 21 — 21.072 x 20 
21 


= - 5.235 tons. 


These complete the web stresses, and leave only those for 
the chords [except (1) and (2) already found] to be deter¬ 
mined. As the counter c x comes into action with the moving 
load over the whole of the bridge, the two upper chord panels 






152 


SWING BRIDGES. 


a and b will sustain the same stress, and panel point 2 at 
the foot of P 3 will be the center of moments for their stress. 
Now if each panel load be called unity, the reactions R x of 
the loads at panel points I, 2, 3, and 4 will be 0.7215, 
0.4627, 0.2431, and 0.0823 respectively; and if each of these 
reactions multiplied by its distance from the center of 
moments be greater than the product of the panel load 
that produces it multiplied by its lever arm, the effect of 
placing that load on the span will be the production of a 
stress of the same kind as that due to the reaction alone. 
If the moments of these unit reactions and that of the unit 
load at panel point 1 be taken about panel point 2, there will 
result : 

(0.7215 x 2 = 1.443) 1 — + 0.4430 

0.4627 x 2 = + 0.9254 

0.2431 X 2 +O.4862 

0.0823 X 2 = + O.I646 

Since all these unit moments are positive, all loads will 
produce compression in a and b. It is clear, also, that the 
heaviest load must be placed at the foot of P 3 ; this will 
locate the two heaviest concentrations at the feet of P z and T lt 
with the head of the train toward P x . The actual moments 
of the panel loads about the foot of P s can now readily be 
found by multiplying those loads by the unit moments given 
above, and then by multiplying the sum of the results by the 
panel length, as follows : 

19.35 x 0.4430 = 8.572 
25.55 x 0.9254 = 23.644 
17.75 x 0.4862 = 8.630 
25.55 x 0.1646 = 4.206 

Total moment . . . 45.052 x 27.5. 

Hence the stress in a and b will be : 

la) = (b) — - 45-052 x 27-5 

29.333 

x sec a — — 45.052 x tan T A x sec a = — 42.31 tons. 


For panel load 1, 

2, 

3 , 

4 , 


11 

u 

u 


u 

u 

{< 


u 

u 

u 




ENDS SIMPLY SUPPORTED. 


153 


In order to find the greatest stress in lower panel 3, 
moments must be taken about the top of P % , By taking 
unit moments, as before, about that center: 

For panel point 1, (0.7215 x 3 = 2.1645) — 2 = 4- 0.1645 

“ “ “ 2, (0.4627 x 3 — 1.3881) —1 = 4- 0.3881 

“ “ “ 3, 0.2431 x 3 = 4-0.7293 

“ “ “ 4, 0.0823 x 3 =4- 0.2469 


Since all these results are positive, all loads on the arm will 
produce tension in lower panel 3, or, what is the same, com¬ 
pression in upper panel c. For the greatest stress in panel 
3, one of the heaviest concentrations must be placed at the 
foot of P<2y so that the two heaviest concentrations will be 
located at the feet of P 2 and T y0 with the head of the train 
toward EP. Hence : 

25.55 x 0.1645 = 4.203 
17.75 x 0.3881 = 6.889 
25.55 x 0.7293 = 18.633 
19.35 x 0.2469 = 4.778 

Total moment . . . 34.503 x 27.5. 


(3) = 34-503 


X 


27-5 

30.67 


= 34.503 x tan F z = 4- 30.95 tons. 


Also : (c) = — (3) x sec a = — 30.985 tons. 

Unit moments about the top of P x for the stress in lower 
panels 4 and 5 will result as follows : 

For panel point 1, (0.7215 x 4 = 2.8860) — 3 = — 0.1140 

“ “ “ 2, (0.4627 x 4 = 1.8508) — 2 = — 0.1492 

“ “ “ 3, (0.2431 x 4 = 0.9724) — 1 = — 0.0276 

“ “ “ 4, 0.0823 x 4 = 4 - 0.3292 


These results show that the panel load at the foot of T x is 
the only one which produces tension in panels 4 and 5 5 
while it is evident that one of the heaviest concentrations 
should be placed at the foot of T x . Hence, by placing the 




154 


SWING BRIDGES. 


advance load of 5.7 tons at the foot of T 2l and the heavy 
load of 25.55 tons at the foot of T x : 

For panel 3, 5.7 x (— 0.0276) — — 0.1575 

“ “ 4, 25.55 x 0.3292 = + 8.411 

Total moment . . . 8.2535 x 27.5. 

Hence: 


(4) = (5) = 8-2535 x 


27-5 

32 


= 8.2535 x tan P x = 4- 7.09 tons. 


In order to find the greatest compression in panels 4 and 5, 
the advance load of 5.7 tons is to be placed at the foot of 
T 2 with the head of the train toward P v Hence: 

For panel 1, 17.75 x (— 0.1140) = — 2.024 

“ “ 2, 25.55 x (-O.1492) = - 3.812 

“ “ 3> 57 x (- 0.0276) = - 0.157 

Total moment . . . 5.993 x 27.5. 

Hence: (4) = (5)= — 5.993 x tan P x = — 5.15 tons. 

The advance load might be so small that it would be neces¬ 
sary to place it at the foot of T x for the greatest compression 
in 4 and 5. 

In order to determine the greatest stresses in panels e and 
6, let unit moments be taken about the foot of P x : 

For panel 1, (0.7215 x 5 = 3-6075) - 4 = — 0.3925 

“ “ 2, (0.4627 x 5 = 2.3135) - 3 = - 0.6865 

“ “ 3> (0.2431 x 5 = 1.2155) - 2 = -0.7845 

“ “ 4> (0.0823 x 5 = 0.4115) — I = — 0.5885 

These results show that all loads produce tension in d and 
e and compression in 6. Hence, by placing the two heaviest 
concentrations at the feet of T x and P % with the head of the 
train toward P x : 





ENDS SIMPLY SUPPORTED. 


155 


- 19-35 x 0.3925 = - 7-595 

- 25.55 x 0.6865 = - 17.540 

- 17-75 x 0.7845 = - 13.925 

- 25.55 x 0.5885 = - 15.036 

- 54096 

2 7 £ 

(6) = — (e) — — 54.096 x - = — 35.42 tons. 

42 

(d — (e) sec ft — 4-35.41 x 1.064 = 4- 37.685 tons. 

(cP) = — (e) tan ft = — 35.41 x 0.364 = — 12.895 tons. 

These complete all the moving load stresses, and, with 
those due to the fixed load, they enable all the resultant 
maxima stresses to be at once written. The following tabu¬ 
lation shows the results of all the computations : 

TABLE I. 


Member. 

Fixed Load. 


Moving 

Load on 

One Arm. 

Both Arms. 



Ti 

+ 3.325 


+ 25-57 

+ 25-57 

T, 

+ 50.02 


+ 38.16 

+ 48.64 

t 3 

+ 36.815 


+ 18.85 
— 14-45 

+ 25 - 3 I 5 

T 4 

+ 22.57 


+ 5.96 
- 34 -oS 

+ 7-51 

7io 

— 8.14 


+ 25-57 
+ 33.21 
— 63.23 

+ 25-55 
+ 20.830 
-48.23 

EP 

Pi 

Pi 

Ps 

cP 

+ 7-00 

— 23-73 
- 29-595 

— 19.09 

- 36.785 


- 59-57 
- 16.57 
- 5-25 

-56.875 

— 18.025 

— 5-235 

— 12.895 


— 62.675 

a 

+ 4.92 


-42.31 

b 

+ 20.37 


— 62.675 

-42.31 

c 

+ 44-98 


— 60.61 

-30.9S5 

V 

+ 98.93 



+ 37 - 6 S 5 

Cl 


+ 92.98 

+ 35-42 

C 



I 

- 4 - 9 i 


+ 44-345 • 

+ 33-83 

2 

-20.35 


+ 44-345 

+ 33-83 

3 

- 44-93 


+ 60.535 

+ 30-95 

4 

- 77-52 


+ 38.79 

j + 7-09 

1 - 5 -i 5 

5 

- 77-52 


+ 38.79 

\ + 7-09 
l - 5 -i 5 

A 

— 92.98 



- 35-41 

u 






















156 


SWING BRIDGES . 


The stresses in Table I show that the tension member 7 i„ 
must be counterbraced so as to sustain the tension of 19.48 
tons due to the moving load on one arm and the fixed load 
compression of 8.14 tons. T 3 is shown under a compression 
of 14.45 tons, due to moving load on one arm, but the fixed 
load tension of 36.815 tons is more than two and one-half 
times as great; hence no counterbracing would be necessary 
in any case. Although 34.08 tons compression is shown 
opposite it belongs to the case in which c 1 does not exist, 
and which will be treated at the end of this Article; it there¬ 
fore needs no further consideration here. The entire lower 
chord must be counterbraced, as must also the upper chord 
panels a, b , and c. These latter members are always built 
with sections for compression. Table II shows the final 
resultant stresses desired. 

TABLE II. 


Member. 

Tension. 

Compression. 

Member. 

Tension. 

Compression. 

T L 

+ 33-895 


a 

+ 4-92 

- 57-755 

Tl 

+ 98 66 


b 

+ 20.37 

- 42.305 

T 3 

+ 62.13 


c 

+ 44-98 

~ 15-63 

Ti 

+ 30.oS 


d 

+1^6.615 


r, 0 

+ 17-43 

— 8.14 

e 

+ 128.39 


Cl 

+ 33-21 


1 

+ 39-435 

- 4.91 

EP 

+ 7.00 

- 56.23 

2 

+ 23.995 

~ 20.35 

Pi 


- 83.30 

3 

+ 15-605 

- 44-93 

P 2 


- 47.62 

A 


— 82.67 





P 3 


- 24.34 

e . 


— 82.67 





cP 


— 49.68 

6 


— 128.39 






It is noticeable in Table I that P 1 receives greater stress 
with moving load on one arm only, than on both arms. This 
is in consequence of the inclination of the upper chord panel 
d, by which it takes a considerable shear that would other¬ 
wise exist in P lt and which would correspondingly augment its 
stress. The same Table also shows that the greatest stresses 
exist in the various portions of the arm under the conditions 
of loading indicated in a former portion of this Article. 

The preceding constitutes all that pertains to the computa- 



































ENDS SIMPLY SUPPORTED . 


157 


tion of stresses in the trusses, but it is still necessary to com¬ 
pute the reactions R 1 and R„ at the extremity of the arm 
and over the center pier respectively. The supports at the 
latching and locking points must be designed to resist the 
reaction R 1} and the drum must be designed to support 
the reaction R a . R 1 will manifestly have its greatest value 
with the moving load on one arm only. The determination 
by trial of the maximum reaction at the foot of EP, including 
the proportionate part of those wheel concentrations in the 
panel adjacent, is somewhat laborious, and as each locomo¬ 
tive and tender weight divided by its total length averages 
3,152 pounds, or 1.576 tons, per lineal foot, it will be essen¬ 
tially accurate and much simpler to assume that the total 
maximum reaction to be supported at the locking and latch¬ 
ing point at the foot of EP is: 

„ 137-5 x 0.775 - 0 , 

R x = - — - — + 6 = 59.28 tons. 

The concentration of 6 tons is added to provide for the 
driving-wheel concentration that must be supported at the 
instant of entering on or leaving the arm while the whole of 
the latter is otherwise loaded. 

The reaction for both arms covered with moving load is 
best determined in precisely the same manner. Each panel 
load will be 27.5 x 0.775 = 21.312 tons. Now the reaction 
R x for four such panel loads will be : 

(.7215 + .4627 + .2431 + .0823) x 21.312 = 1.5096 x 21.312. 

.*. R x = 32.173 tons. 

« 

Hence: 

RJ = 4 x 21.312 - 32.173 + x 21.312 + 6 = 77.674 tons. 

2 7-5 

The panel load added is for the half panel adjacent to the 
center and the central panel, while the 6 tons is for the driv¬ 
ing-wheel concentration, as just explained in connection 
with R v To R t ' must be added the total fixed load carried 




158 


SWING BRIDGES. 


to the center. From the data given at the beginning of this 
Article, that total fixed load will be: 

24 x 0.85 

4 x 11.7 + 5 H——-- = 62.0 tons. 

2 


Hence the total reaction R 2 to be supported at the drum 
will be: 


R . 2 = 77.674 + 62 — 139.674 tons. 


As the ends of the span are simply supported, no fixed 
load reaction is to be added to the value of R x as determined 
by the moving load only. 

The omission of counters. 

It has already been remarked that the best design, from a 
purely engineering point of view, is secured by omitting the 
counters and counterbracing the main web members. In the 
present instance the counter c x would be omitted and the 
main members T A and P 3 counterbraced. The member T 10 
will also receive much greater tension than in the preceding 
case. 

In considering these counter stresses, the arm of the 
bridge is to be treated as a simple non-continuous span, as 
was done in the earlier part of the Article, with the moving 
load passing from the center toward the end. Equation (4) 
of Article 20 shows that the third driver of the front locomo¬ 
tive should be at the foot of T l0 in order to give that mem¬ 
ber its greatest tension. The first parenthesis of the second 
member of that equation disappears in this case because m 
is zero. The resulting reaction at the foot of EP is 50.86 
tons, of which 5.70 tons is due to the wheel concentrations 
between the foot of EP and the foot of T lQ . Hence, by 
moments about the chord intersection : 



(50.86 — 5.70) 20 

21 


= + 43.01 tons. 


The moving load compression in T 4 has already been found 
to be 34.08 tons, in deciding upon the necessity for c x in the 




ends simply supported . 


l S 9 


t 


preceding case. The same position of the moving load (the 
third driver at the foot of P^j will give the greatest tension 

in P 3 , and using the data already employed for T iy there will 
result: 

n 32.1 x 20 — 5.72 x 21 
P * =--- + 23.73 tons. 

Again, it was found that the greatest moving load compres¬ 
sion in T s would be 14.45 tons, while the fixed load tension is 
36.815 tons ; hence T s can never suffer compression. The re¬ 
sults for T 3 show also that the fixed load compression in P 2 will 
always largely overbalance any possible moving load tension. 
The only counterbracing needed, therefore, will be that for 
the members T 4 and P 3 . These computations, in combina¬ 
tion with the results given in Table I, show that T 10 , T 4 , and 
P 3 will sustain the following resultant stresses : 

T l0 ; + 34.87 tons. — 8.14 tons. 

Ti ; + 30.08 “ — 11.51 “ 

P 3 ; + 4-^4 “ - 24.34 “ 

All the other stresses remain unchanged. 

Art. 37.—Ends Simply Supported—Two Points of Support at Center— 

Complete Continuity—Example. 

The case of complete continuity to be considered in this 
Article involves the existence of web members in the panel 
over the drum {i.e., the span / 2 of Article 35), designed by 
actual computation to take the shear which may pass the 
center when the structure carries unbalanced loads. 

It has been shown, in Article 35, that when the condition 
of continuity is fulfilled, which is the only condition there 
contemplated, either of the reactions R 2 or R 3 may be nega¬ 
tive, i.e., downward, while both the reactions R x and are 
always positive, or upward. It will be found that provision 
of the nature of heavy anchorage must be made in order to 
meet the requirements of R 2 and R 3 with some conditions of 
loading of draw spans. This indicates that the web members 
in the center panel will be very heavy, and such will be found 
to be the case. 



i6o 


SWING BRIDGES. 


The moving load stresses in all cases will be determined 
by means of the formulae of Article 35. The truss to be con¬ 
sidered is the same as treated in the preceding Article, and 
is shown by Figure 1 of Plate X. The data, reproduced 
from Article 36, are as follows: 

Length between centers of end pins . . 295.5 feet. 

Panel length in arms.27.5 

Length of center panel or span ... 20 


Center depth at CP — 42 feet. Depth at T x = 32 feet. 
Depth at P 2 = 30! “ “ “ P 3 = 29J “ 

“ “ T l0 — 28 “ Truss centers 16 ft. apart. 


Weight of rails, ties, and guards, with bolts and connec¬ 
tions, 0.2 ton per lineal foot. 

Total lower panel fixed load .... 8.325 tons. 

“ upper “ “ “ . 3.375 “ 

Total ..11.700 “ 


Lower panel load at end of arm ... 5 tons. 

Upper “ “ “ top “ CP .... 2.94 “ 

The trigonometric quantities required will not be re¬ 
peated. 

As the ends of the span are simply supported, the fixed 
load stresses are those existing when the draw is open, and, 
hence, are identical with those found for the fixed load in Table 



Fig. 1. 


I of the preceding Article. As they are there arranged in 
convenient shape for combination with the moving load 
stresses about to be found, they will not be reproduced here. 













ENDS SIMPLY SUPPORTED. 


161 


The moving load is identical with that used in Article 36, 
and is shown by the diagram on preceding page. 

For the same reasons given in that Article, the system of 
panel concentrations which was used there will be employed 
here, instead of the wheel weights shown in the diagram. 
These panel concentrations are as follows: 


Advance load 
At panel 2 . 


5.7 tons. At panel 1 
1775 “ “ “ 3 

19-35 “ 


25.55 tons. 
25.55 “ 


On account of the complete continuity at the center, there 
will be no condition of a simple, non-continuous span for the 
moving load on one arm only, but all moving load stresses 
will be found by the use of the preceding panel concentra¬ 
tions in connection with the formulae of Article 35. The 
reactions R x and R x will be precisely the same as those found 
on page 146 of Article 36. They are: 

Panel load P at panel I, R x = 0.716102 P; R 4 = 0.005443 P. 

“ “ “ “ 2, R x = 0.453178 P; R± — 0.009526 P. 

“ “ “ “ 3, R x = 0.232204 P ; Ri = 0.010886 P. 

“ “ “ “ 4, R x = 0.074153 P\ R a = 0.008165 P 


As unbalanced loads will be considered in this case, these 
reactions will frequently be used separately in the succeeding 
computations; but in those special circumstances which cause 
the loads to be balanced, they will be united by addition, as 
was done in the preceding Article. 

As all loads produce positive or upward reactions at each 
end of the bridge, the post EP will receive its greatest stress 
when the moving load completely covers both arms, and in 
such a manner that the reaction at its foot will be the great¬ 
est possible, as shown by the values of the panel reactions 
given above. The reaction will therefore be: 

25.55 x 0.7161 = 18.297 tons. 

17.75 x 0.4532 — 8.044 “ 

25.55 x 0.2322 = 5.932 “ 

19.35 x 0.0741 = 1.434 “ 


Left arm total = 337°7 


u 



162 


WING BRIDGES. 


19.35 x 0.0054 = 0.104 ton. 

25.55 x 0.0095 = 0.243 “ 
17.75 x 0.0109 = 0.194 “ 

25.55 x 0.0082 = 0.209 “ 


Right arm total = 0.750 “ 


{EP) - - (33.707 + 0.75) sec EP= — 34-457 x 1.4 

= — 48.24 tons. 

(1) = (2) = 34-457 x tan EP = + 33-835 tons. 


As T k cannot sustain any compression, the hanger T t 0 will 
be stressed by the entire load at its foot, unless that load 
added to the coexistent vertical shear in the upper chord 
panel a exceeds the reaction at the foot of EP. Since the 
latter has been shown to be 34.457 tons, while the maximum 
load at the foot of P lQ is only 25.55 tons, that excess cannot 
exist, but c{ will be required to carry over shear to make up 
the reaction found. Hence: 


(Tjo) = + 25.55 tons. 

Since the function of c x is to carry over shear to make up 
the reaction at the foot of EP, its stress will be the greatest 
when that reaction exceeds by the greatest amount possible 
the moving load between it and c lt added to the coexistent 
shear in upper panel a. As the advance load is small in com¬ 
parison to that which follows it, this condition will exist when 
the moving load covers both arms, with the advance load at 
the foot of T 10 . The reaction at the foot of EP will then be: 


5.7 x 0.7161 = 4.082 tons. 

25.55 x 0.4532 = 11.579 “ 

17.75 x 0.2322 = 4.122 “ 

25.55 x 0.0741 = 1.893 “ 

Total from left arm .... 21.676 “ 

Right arm reaction as above. 0.75 “ 

Total reaction .... 22.426 “ 



22.426 x 20 — 5.7 x 21 


sec c x = + 20.925 tons. 


22 






ENDS SIMPLY SUPPORTED . 163 

The advance load would have to be placed at the foot of 
P 3 if it were sufficiently large. 

At the end of the Article the case will be considesed in 
which the counter c x is omitted and the members T 4 and P s 
counterbraced. It will then be shown that T s and P 2 can 
never sustain any other kinds of stresses than those induced 
by the fixed load. Hence, no other counter than c x will be 
required in the present instance, and the main web stresses 
will next be found. 

Since the function of the main web members is to carry 
load or shear over to the center, one condition to be fulfilled in 
order that they may receive their greatest stresses is, in gen¬ 
eral, to make the reaction at the end of the arm in which they 
are located as small as possible. As the main web stresses in 
the left arm are now sought, therefore, no moving load must 
be placed o?i the right arm ; because all such load increases 
the upward reaction at the foot of EP. 

This condition holds in general because the chords of draw 
spans are usually either parallel or so inclined to each other 
that their intersections lie without the span. If, however, their 
intersections lie within the span, the reactions at the free end 
of the span will have moments of the same sign as those loads 
between the moment origin and the web members in question. 
Hence in those cases, the reactions (with moving loads omitted 
between moment origin and the free end) should be as large 
as possible, and the moving load should cover the whole of 
the other arm. 

The intersection of the panels d and 5 is 0.8 panel length 
from the foot of EP , and 0.2 panel length from the foot of T 10 ; 
hence the entire bridge is to be covered with moving load 
for the greatest stress in P u precisely as was done in the pre¬ 
ceding Article for the same member. Therefore, from that 
Article: 

(Pi) = - 56.875 tons. 

The member T x is again a simple hanger, and its greatest 
stress is the greatest reaction at its foot. Hence: 

T x = + 25.57 tons * 


164 


SWING BRIDGES. 


For the remaining web members the moment origin is 
twenty panel lengths from the end of the span, and the mo¬ 
ments of the reactions will have signs opposite to those of 
the loads. No moving load must, therefore, be placed on 
the right arm, in order that the reactions may be as small 
as possible. 

The advance load will be placed at the foot of T x for the 
greatest stress in thus making the reaction at the foot 
of EP : 


For panel 1, 25.55 x 0.7161 = 18.295 tons. 
“ “ 2, 17.75 x 0.4532 = 8.04 

“ “ 3* 25.55x0.2322= 5.933 “ 

“ “ 4, 5.7 x 0.0741 = 0.422 “ 



Total reaction = 32.690 “ 


25.55 x 21 + 17.75 x 22 4- 25.55 x 23 — 32.69 x 20 

23 


sec T 2 


= + 49.37 tons. 


The greatest stress in T 3 requires the advance load at the 
foot of T 2 , giving the reaction : 

For panel 1, 17.75 x 0.7161 = 12.71 tons. 

“ “ 2, 25.55x0.4532=11.58 

“ “ 3, 5.7 X 0.2322 = 1.325 “ 

Total reaction = 25.615 “ 


( *r\ x 21 + 25.55 x 22 - 25.615 x 20 

( 1 3) =-—- sec T 3 

= + 25.79. 

Since P 2 takes its greatest stress with T 3 : 



17.75 x 21 + 25.55 x 22 — 25.615 x 20 


23 


= — 18.37 tons. 







ENDS SIMPLY SUPPORTED . 165 

The greatest stress in 7 ^ requires the advance load to be 
placed at the foot of T 3 . Hence the reaction will be: 


For panel 1, 25.55 x 0.7161 = 18.295 tons. 
“ “ 2, 5.7 x 0.4532 = 2.585 “ 

Total reaction = 20.880 “ 


(T<) 


25.55 x 21 — 20.88 x 20 


21 


sec Tt= + 7.765 tons. 


Since P 3 takes it greatest stress with the same position of 
moving load : 



25.55 x 21 — 20.88 x 20 
22 


— 5.405 tons. 


In order to illustrate the effect of moving the load so as 
to bring the advance concentration of 5.7 tons at the foot of 
the diagonal whose stress is sought, let the stress in T 3 be 
found in that manner. The reaction as found for T 4 will be 
20.88 tons. Hence: 


25.55 x 21 + 5.7 x 22 — 20.88 x 20 _ 

( 71 )' = - sec T z 

= + 14.91 tons. 

This result is only a little more than one-half the greatest 
stress in T 3 as found above. If the advance load had been 
large, the result would have been different. 

The greatest stresses in the members cc and LS in the center 
panel are next to be found. These members carry the shear 
past the center for all unbalanced loads on either arm ; hence 
the stresses in them will occur when there exists the greatest 
amount of unbalanced load— i.e., when the moving load covers 
one arm while the other arm is free of it. This also follows 
clearly from the fact, demonstrated in Article 35, that all 
load in the arm, or span, l x causes R 3 to be negative and R 2 
positive; and, hence, since the stresses in cc and LS result 
from the downward pull of R 3 , any load in the arm, or 






SWING BRIDGES. 


166 

span, / 3 would reduce the negative R 3 by balancing some load 
in l lf and thus correspondingly reduce the stresses in question. 
Those considerations are concomitant with complete conti¬ 
nuity over the center. 

On page 161 are found values of R x and R A for a unit panel 
load at each panel point of either arm, and by using these in 
Equation (ya) of Article 35 the following values of R 3 result, 
remembering that P is unity, that c = 6.707, and that n has 
the values 0.2, 0.4, 0.6, and 0.8 respectively: 

For panel 1, R 3 = — 0.6044 

“ “ 2, ^=-1.0579 

“ “ 3. ^3 = ~ 1.2095 

“ “ 4. ^3 = — 0.9076 

The shear in the center panel will be R 3 4- R 4 (a numerical 
difference); hence, by using actual panel loads: 

(- 0.6044 + 0.0054) x 19-35 = — n.59 tons. 

(- 1.0579 + 0.0095) x 25.55 = - 26.785 “ 

(— 1.2095 + O.OIO9) X I7.75 = — 21.275 “ 

(— 0.9076 4- 0.0082) x 25.55 = — 22.98 “ 

.*. R 3 4 - Ri = — 82.630 “ 

Since the two rods cc pull against opposite ends of the strut 
LS, the preceding shear of 82.63 tons will be equally divided 
between them. Hence: 

(cc) = 41.315 x sec cc = 4- 57.715 tons. 

The stress in LS is of course the horizontal component of 
that in cc ; therefore : 

(LS) — — 41.315 x tan cc — — 40.33 tons.' 

The preceding calculations show, therefore, that the total 
downward reaction to be supplied, or upward pull to be re¬ 
sisted, at the foot of each cP is 82.63 4- 0.75 = 83.38 tons; 
and there must be sufficient weight of drum, cross beams, 
etc., as well as strength of connections, to fulfill the require¬ 
ments of that condition. 



ENDS SIMPLY SUPPORTED. 


167 


Since the center panel, or span, is divided into two stories, 
the upper half of ^will sustain a stress different from that 
in the lower half in consequence of the action of the two 
parallel tension braces cc. Again, unbalanced moving loads 
will produce stresses in one of the cP posts that are not only 
different in amount from those in the other, but, also, different 
in kind. 

It has just been shown that the cc tension braces inclined 
in the same direction as P l receive their greatest stresses 
when the left arm is entirely covered with moving load, the 
right arm being free from it. It necessarily follows that the 
left post cP receives at the same time its greatest compres¬ 
sion. The same condition of loading, it has also been shown, 
induces the greatest possible downward reaction, or pull, at 
the foot of the right cP. In fact, this pull produces a heavy 
tension in the lower story of that cP at the foot of which the 
reaction R 3 exists; but in the upper story of the same member 
a small tension only will be found, due to the small moving 
load compression in the right-hand member d induced by the 
reaction R±. It will be further apparent that the moving load 
on one arm only will cause the greatest compression in that 
^nearest the loaded arm, if it be observed that the member 
d takes its greatest tension with the same condition of load¬ 
ing; because if the other arm be loaded, the upward reaction 
at the end of the arm will be increased by the amount of R i} 
which will correspondingly relieve the tension in d. 

Let the greatest compression in the lower story of cP be 
found with the same condition of moving load used for cc, and 
in doing this let the truss be supposed to be divided through 
d, cP, and 6 (the cc which is divided does not act and is there¬ 
fore neglected). Moments will be taken about the intersec¬ 
tion of d and 6 in the left arm, by the reactions R± and R z . 
Hence, remembering that 20.5 27.5 = 0.7455 : 

Lower half (cP) = °-75 x 9 . 9 455 - 83.38 x 4-9455 

dL • Jm 


— — 96.4 tons. 



168 


SWING BRIDGES. 


This is, of course, equal to the vertical component of the 
tensile stress in d added to the vertical components of the two 
stresses (cc). It will presently be shown that the tension in d 
is 37.875 tons; hence: 

Lower half (cP) = 37.875 x 0.364 + 2 x 41.315 = — 96.4 tons. 

The compression in the upper story of cP will be simply 
that in the lower story diminished by the vertical component 
of (cc). Hence: 

Upper half (cP) — — (96.4 — 41.315) = — 55.085 tons. 

This can, again, be checked by moments by supposing the 
truss divided through d, upper half cP } LS, cc , and 6, and by 
including the moments of (LS) and (cc) now known ; but it is 
quite unnecessary. 

Continuing the same position of loading, the tension in 
the lower half of the other cP will now be found. Let the 
truss now be supposed to be divided through the right 
arm d, lower story of right-hand cP, cc , and 6; then let mo¬ 
ments be taken about the intersection of 6 and d in the right 
arm. Remembering that = 0.75 ton, and R s = 83.38 
tons, while the lever arm of (cc) = 57.715 tons is 4.2 panels 
divided by sec for cc : 


0.75 x 0.8 + 83.38 x 4.2 
4.2 


Lower half tension (cP) = 


57.715 x 4.2 
4.2 x sec cc 


= + 42.21 tons. 


This result can be checked by adding the vertical compo¬ 
nent of the upper (cc) to the vertical component of the com¬ 
pression now existing in d (in consequence of the upward 
reaction R 4 ), whose horizontal component will presently be 
shown to be 2.455. Hence, this checking operation gives : 


41.315 + 2.455 x ian ft = 42.21 tons. 




ENDS SIMPLY SUPPORTED. 169 

The upper half of c/ 5 receives only the vertical component 
of the compression in d; hence : 

Upper half tension {cP) = 2.455 x tan ft = + 0.895 tons. 

This result can also be checked by moments in the same 
general manner indicated for the upper half of the other cP, 
but it is not necessary. 

The computations for the web stresses are thus completed, 
leaving those for the chord stresses yet to be shown. 

Unit panel moments, as described on page 152 of Article 36, 
will first be taken about the foot of P s for all the panel points 
of the two arms, those on the right arm being represented 
by R iy and the following will be the results: 

For right arm, R A x 2 —- + 2R 4 

“ panel 1, 0.7161 x 2 — 1 = + 0.4322 

“ “ 2, 0.4532 x 2 =+ 0.9064 

“ “ 3, 0.2322 x 2 =4- 0.4644 

“ “ 4, 0.0741 x 2 = + 0.1482 

All these results are positive, which shows that all loads 
on both arms produce compression in b ; but with the entire 
left arm loaded, the counter c x comes into action and causes 
the stress in a to be the same as that in b. Therefore, for the 
greatest compression in a and b y moving load must be so 
placed on the right arm as to give i? 4 its greatest value of 
0.75 ton (as already shown), while the left arm is entirely 
covered, with the heavy concentrations at panels 2 and 4. 
The desired moment will then be: 

0.75 x 2 =1.5 

19.35 x 0.4322 = 8.363 
25.55 x 0.9064 = 23.159 
17.75 x 0.4644= 8.243 
25.55 x 0.1482 = 3.786 

Total . . . 45.051 


(a) = (*) = - 


45.051 x 27.5 


see a = — 45.051 


x tan x sec a = — 42.31 tons. 


29-333 




170 


SWING BRIDGES. 


Again, for the chords c and 3, unit moments about the 
foot of P 2 give : 

^4 x 3 + 3*4 

0.7161 x 3 — 2 = + 0.1483 
0.4532 + 3-1=4- 0.3596 
0.2322 x 3 = + 0.6966 

0.0741 x 3 = + 0.2223 

As all these results are positive, it follows that all panel 
loads produce compression in c and tension in 3. It is also 
evident that the heavy concentrations must be placed at 
panels 1 and 3. Introducing the panel concentrations: 

0.75 x 3 = 2.25 

25.55x0.1483= 3.789 
1775 x 0.3596 = 6 .383 
25.55 x 0.6966 = 17.798 
19.35 X 0.2223 = 4.301 

Total . . . 34.521 

( 3 ) = 34 ' ? 3 o 666 7 ' 5 = 34 ' 521 tan Ts ~ + 30,965 tons - 

Also: (c) = — (3) sec a — — 31.005 tons. 

For the stresses in 4 and 5 (equal to each other), the unit 
moments are to be taken about the top of P l ; they are as 
follows : 


For P 4 , 

“ panel 1, 


For Pi, Pi x 4 = + 4P 4 

“ panel 1, 0.7161 x 4 — 3 = — 0.1356 
“ “ 2, 0.4532 x 4 — 2 = — 0.1872 

“ “ 3, 0.2322 x 4 — 1 = — 0.0712 

“ “ 4> 0.0741 x 4 = + 0.2964 

These results show that all loads on the right arm with 
that at panel 4, only, produce tension in 4 and 5. Hence, 
the moving load on the right arm is to remain as before, 
while the heavy concentration is to be placed at the foot of 




ENDS SIMPLY SUPPORTED . iyi 

T lt with the advance load of 5.7 tons at the foot of P 2 . By 
the introduction of these moving loads: 

For 0.75 x 4 = + 3 

“ panel 4, 25.55 x 0.2964 = + 7.573 


+ 10-573 

“ “ 3, — 5.7 x 0.0712 = — .406 

Total ... + 10.167 

(4) = (5) = 10.167 x tan T 2 = 4- 8.735 tons. 

Since the loads at panel points i, 2, and 3 produce com¬ 
pression in 4 and 5, the greatest compression possible must 
be found. It will result from placing the heavy concentra¬ 
tion at panel point 2, with the advance load of 5.7 tons at 
panel point 3. The actual concentrations then give : 

For panel point I, — 17.75 x 0.1356 = — 2.407 
“ “ “ 2, — 25.55 x 0.1872 = — 4.783 

“ “ “ 3, — 5.7 x 0.0712 = — .406 

Total ... = — 7.596 

.*. — (4) = — (5) = — 7.596 x tan T 2 — — 6.525 tons. 

It is thus seen that the panels 4 and 5 are the only por¬ 
tions of the lower chord that can ever be subjected to com¬ 
pression by the moving load. 

In order to determine the stress in d, the unit moments 
must be taken about the foot of cP y and they are : 

For right arm, P 4 x 5 =4- 5^4 

“ panel point I, 0.7161 x 5 — 4 = — 0.4195 
“ “ “ 2, 0.4532 x 5 - 3 = - 0.7340 

“ “ “ 3, 0.2322 x 5 — 2 == — 0.8390 

“ “ “ 4, 0.0741 x 5 — 1 = — 0.6295 

Whence, all loads on the right arm give compression to 
d , and they must be removed when the greatest tension in d 





172 


SWING BRIDGES. 


is sought. The greatest moving load compression in d will 
therefore be: 


27 5 

(d) = — 0.75 x 5 x x sec ft = — 2.455 x sec fi 

4 2 


= — 2.61 tons ; 


and since under this loading (on the right arm only) the rods 
cc sloping upward to the left do not act, the compression 
in e will be the horizontal component in d . Hence: 


(e) = — 2.455 tons. 

If all the load on the right arm is removed, and all that on 
the left arm retained, the preceding show that d will be sub¬ 
jected to its greatest tension. By placing the heavy con¬ 
centrations at panel points 2 and 4, and introducing all the 
panel concentrations on the left arm in the unit moments, 
there will result: 


For panel point 

(( U u 

«( u u 


1, 

2, 

3 > 


u u 


a 


4 > 


19.35 x 0.4195 = 8.118 

25-55 x 0.7340 = 18.753 

17.75 x 0.8390= 14.892 

25.55 x 0.6295 = 16.084 


Total . . . 57.847 


(d) = 57.847 x x sec fi=z 37.875 x sec p 

42 


= +40.3 tons. 


Under this condition of loading, the rods cc sloping up¬ 
ward toward the right do not come into action ; therefore 
the compression in 6 must equal the horizontal component 
of the tension in d\ or: 

( 6 ) = — 37- 8 75 to ns. 

It has been shown that moving load on one arm, only, pro¬ 
duces compression in e ; hence, to obtain the greatest tension 





ENDS SIMPLY SUPPORTED. 


173 


in that member, the whole of both arms must be loaded with 
balanced moving loads, which is precisely the condition used 
in Article 36, on page 155, where there was found: 

(e) — + 35.42 tons. 

This completes the computations for all the moving load 
stresses, and Table I shows them grouped so that they may 
be combined with those due to the fixed load as found in 
Article 36. 

TABLE I. 


Member. 

Stress. 

Member. 

T x 

+ 25.57 tons. 

c 

T\ 

+ 49-37 “ 

d 

t 3 

+ 25-79 “ 

T< 

+ 7-765 “ 



+ 25.55 “ 

c 

Cl 

+ 20.925 “ 

I 

EP 

— 48.24 “ 

2 

Pi 

-56.875 “ 

3 

P 3 

-18.37 “ 


P 3 

- 5-405 “ 

4 

Upper cP 

j -f 0.895 “ 

\ - 55-085 “ 

5 

Lower cP 

j + 42.21 

6 


( — 96.40 

cc 

a 

-42.31 

LS 

b 

— 42.31 



Stress. 


— 31.005 tons. 

+ 40.3 

— 2.61 

+ 35-42 

- 2.455 

+ 33-835 
+ 33-835 

+ 30-965 

j+ 8.735 
{- 6.525 
+ 8.735 

— 6.525 

- 37-875 
+ 57-715 

- 40-33 


By comparison of these results with those of Table I on page 
155, it is observed that the assumption of complete continuity 
produces web stresses either slightly greater in every instance 
(except those due to the moving load on one arm only) than 
the hypothesis of partial continuity, or identical with it. The 
chord stresses are also usually slightly greater, or equal, ex¬ 
cepting, also, those due to moving load on one arm only in 
the case of partial continuity, which are materially greater. 
Toward and at the ends of the span, therefore, the assump¬ 
tion of a simple non-continous span gives much larger stresses, 
both in chords and web, than are found underthe supposition of 
continuity, as would be anticipated. With these exceptions, 
however, the differences are unimportant, and either method 

















174 


SWING BRIDGES . 


could be used with indifference, and that selected which 
would result in the least labor of computation. 

An enormous difference is seen, however, in the magnitude 
of the stresses both for the center posts and central diago¬ 
nals, and in the labor of their computation, cc and LS suffer 
no stress under partial continuity, but are subjected to heavy 
stresses if the continuity is complete. The posts cP also suf¬ 
fer very heavy compression in the latter case, and a compar- 
itively light compression in the former, due only to the fixed 
load. The lower stories of the same members are also sub¬ 
jected to heavy tension when the continuity is complete. In 
the present instance this tension, added to the vertical com¬ 
ponent of the stress in the lower cc, is 42.21 + 41.315 = 83.525 
tons; and since the fixed load compression in cP is, by Table 
I of Article 36, 36.785 tons, the total maximum upward pull 
to be resisted at the foot of cP is : 

83-525 ~ 36785 = 4674 tons. 

Now the lower chord fixed load at the foot of cP is 
24 x 0.3025 = 7.26 tons, and the weight of the drum, etc., 
which may be assumed to be concentrated at the same point, 
is not more than 9 or 10 tons; or, say, 974 tons. Hence the 
amount of anchorage which would have to be provided under 
the assumption of complete continuity is : 

4674 — (7.26 + 974) = 29.74 tons ; 

and unless this were supplied, each foot of cP would, in turn, 
rise and fall with varying conditions of moving load, so that 
not only very destructive hammering would take place, but, 
also, the condition of those spans continuous over two sup¬ 
ports would be displaced by that of two unequal spans con¬ 
tinuous over one support, thus vitiating the computations 
underlying the design of the trusses. 

The realization of the conditions requisite for the case of 
complete continuity, therefore, involves considerably in¬ 
creased difficulty and expense in connection with the design 
of the center portion. 


ENDS SIMPLY SUPPORTED. 


175 


Inasmuch as the main truss members have been found to 
sustain essentially the same stresses in either case, except 
where those under partial continuity largely exceed the others, 
and inasmuch as all draw-bridge formulae drawn from the 
theory of the continuous beam involve the uniformity of 
the moment of inertia of all normal sections, thereby incur¬ 
ring a very considerable error in at least some of the resulting 
computations, it is much more rational and in harmony with 
better judgment to use the methods of partial continuity in 
all cases. These and similar considerations have led engi¬ 
neers to almost universally adopt the partial-continuity as¬ 
sumption in the construction of swing bridges. 

It is to be observed that the unbalanced uplift at the foot 
of cP and the stresses in cc and LS will increase rapidly as the 
length of the center panel or span decreases; and, hence, that 
it is advisable to make that center panel as long as possible. 
The method of computation will be in no way changed if 
the center panel is designed with one story only, instead of 
two. 

By combining the fixed load stresses in Table I of Article 
36 with those due to the moving load given in Table I of this 
Article, Table II, giving the resultant stresses in all the mem¬ 
bers, at once results. 


TABLE II 


Member. 

Tension. 

Compression 

T 1 

Tons 
+ 33-895 

Tons 

T 

+ 99-39 


T 3 

+ 62.605 


T 

+ 30.335 

— 8.14 

To 

+ 17.41 

C\ 

+ 20.925 


EP 

+ 7-0 

- 41.24 

Pi 


— 80.605 

P a 


- 47 - 9^5 

Ps 


- 24.495 

Upper cP 


- 9 1 *87 

Lower cP 

+ 5-425 

- 133-85 

a 

+ 4.92 

- 37-39 


Member. 

Tension. 

Compression. 


Tons 

Tons 

b 

+ 20.37 

- 21.94 

c 

4 - 44-98 


d 

+ 139-23 


e 

4- 128.40 


1 

-1- 28.925 

- 4.9I 

2 

+ I 3-485 

- 20.35 

3 


- 44-93 

4 


— 84.045 

5 


- 84045 

6 


- 130.855 

cc 

+ 57-715 


LS 


- 40-33 





































176 


SWING BRIDGES. 


These resultant stresses are not very different from those in 
Table II of page 156, except in those members already indi¬ 
cated. In this case, upper chord c and lower chord 3 do not 
need counterbracing, while they required such treatment in 
the case of partial continuity. In actual practice, however, 
the entire lower chord, and the upper chord panels a , b , and c , 
would all be designed with compression cross sections— i.e., 
they would be counterbraced. The lower story of cP must 
also be counterbraced, and ordinary construction would make 
it so even if it were not required. 

The existence of the central diagonals cc designed to trans¬ 
fer shear, results in some ambiguity in the upper chord stresses 
at the center. As the moving load passes on the bridge, one 
pair of those diagonals receive their greatest stresses, and 
then they may either be supposed to be relieved with the 
further progress of the train, or they may be supposed to hold 
essentially their greatest stresses while the other pair gradu¬ 
ally take the same condition, as the moving load covers the 
entire structure. Either supposition fulfills the requisites for 
equilibrium, but the former will give the greatest possible 
stress in e and was used in the preceding computations, be¬ 
cause the stress in the same upper chord member will be 
decreased under the latter supposition by an amount corre¬ 
sponding to the horizontal components of the stresses in the 
diagonals cc. This condition of ambiguity cannot be avoided 
under the assumption of complete continuity, but disappears 
if the continuity is assumed to be partial only. The proper 
method, therefore, is to compute the greatest possible stress 
in each member, and use it in the design, and this has been 
done in the present case. 

The reaction R x and R 2 at the ends and adjacent to the 
center, respectively, remain to be written, but they can be 
taken directly from results already found in the preceding 
Article. For this purpose, and for the reasons fully explained 
in Article 36, the moving load will be taken as uniform and 
at 0.775 ton P er lineal foot for each truss. The greatest end 
reaction will exist with the moving load over the entire struct¬ 
ure, and it has been shown on page 157 that its amount will be : 


ENDS SIMPLY SUPPORTED. 


1 77 


R\ = 32.173 + 


21.312 

2 


+ 6 = 48.829 tons. 


The half panel load is that adjacent to the end, which 
does not affect the trusses but forms a part of the reaction 
to be supported by the truss ends, while the 6 tons is the 
wheel concentration. Since the moving load on each arm 
produces a negative or downward reaction at the opposite 
side of the drum, the greatest reaction R 2 will be produced 
with the moving load on the adjacent arm only. The reac¬ 
tion R x due to the four panel loads at the panel points 1, 2, 3, 
and 4 will be: 

R x = (0.7161 + 0.4532 + 0.2322 + 0.0741) = 31.448 tons. 


Hence: 


R-2 = 4 x 21.312 — 31.448 + 


24 

27.5 


x 21.312 + 6= 78.399tons. 


The third term in the second member is for the half panel 
adjacent to the center and the central panel, while the 6 
tons is for the driving-wheel concentration, as already ex¬ 
plained. As shown on page 158, the total fixed load at the 
foot of P 1 is 62 tons; hence the total reaction desired is: 

R 2 = 78.399 + 62 = 140.399 tons. 


The omission of counters. 

If all counters are omitted, it will usually be necessary 
to counterbrace some of the main web members. In the 
present case, T 4 will be most in need of such treatment. The 
position of moving load required to give its maximum com¬ 
pression to T 4 is the same as that used in finding the great¬ 
est tension in c x on page 162, and the corresponding reaction 
at the foot of EP was there found to be 22.426 tons. 

Hence: 



22.426 x 20 — 5.7 x 
21 


21 


sec T i = — 21.47 tons. 


12 





i ;8 


SWING BRIDGES. 


But it has been shown that the fixed load tension in 7 " 4 is 
22.57 tons. No counterbracing, therefore, according to these 
results, is required. But the tension excess is so small that 
the member should be made with a compression section, 
otherwise the moving load shock might overcome the excess 
and cause the member to buckle. A rule of good practice is 
to counterbrace if at least one-fourth of the computed mov¬ 
ing load stress added to itself overcomes that due to the 
fixed load, and perhaps a better rule is to add one-third. 

With the same position of loading there will result: 



22.426 x 20 — 5.7 x 21 
22 


= + 14.95 tons. 


If one-fourth of 14.95 tons be added to itself, the result 
will be less than 19.09 tons, the fixed load compression in P 3 ; 
hence no reversion of stress can take place, although the 
character of the pin connection would enable very consider¬ 
able tension to be resisted. 

For the greatest compression in T 3 , and tension in P 2f the 
advance load of 5.7 tons must be placed at the foot of P 3 . 
The reaction at the foot of EP will then be : 


5.7 x 0.4532 = 2.583 
25.55 x 0.2322 = 5.932 
17.75 x 0.0741 = 1.316 

Total from left arm . . . 9.831 tons. 
Right arm reaction . . . 0.75 “ 

Total . . . 10.581 


Hence: 



10.581 x 20 — 5.7 x 22 
22 


sec T 3 = — 5.265 tons- 


And: 



10.581 x 20 — 5.7 x 22 


+ 3.75 tons. 


23 







TABLES AND DIAGRAMS. 


179 


Both of these results are so small in comparison with the 
opposite fixed load stresses that no counterbracing is re¬ 
quired. 

These computations show that even if the counter c x is 
omitted, no counterbracing will be. required, except in the 
case of 7*4, as explained. 


Article 37 a.—Tables and Diagrams.—Turntables and Engines. 

The labor of stress computations by the methods given in 
the three preceding Articles can be reduced to a very small 
amount by the aid of the tables and diagrams which follow. 
They are devised for the purpose of showing the reactions at 

TABLE I. 




Rx 



R , 


n = 7 

c 

c 

c 

c 

c 

c 


8.822 

6.643 

4-395 

8.822 

6.643 

4-395 

0 

I. 

I. 

I. 

0.0 

0.0 

0.0 

•05 

0 92754 

0.92823 

O.92951 

.001143 

.001424 

.001899 

. 10 

0.8554 

0.85679 

0.85933 

.002269 

.002826 

.00377 

.15 

0.78398 

0.78601 

O.78976 

.003360 

.004186 

.005583 

.20 

0-71355 

0.71621 

0.72112 

.004401 

.0054S2 

.007311 

• 25 

0.64447 

0.64771 

O.6537I 

.005372 

.006692 

.008926 

♦ 30 

0.57708 

0.58085 

O.58784 

.006257 

•007794 

.010396 

•35 

0.51171 

0.51596 

O.52382 

.007039 

.008768 

.011694 

.40 

0.44871 

0.45336 

O.46195 

.007701 

•009593 

.012795 

• 45 

0.38841 

0.39333 

O.40256 

.008226 

.010247 

.013667 

•50 

0.33115 

0.33634 

0-34593 

.008595 

.010706 

.01428 

•55 

0.27727 

0.28257 

O.29239 

.00S792 

.010952 

.014607 

.60 

0.2271 

0.23241 

O.24223 

.008801 

.010963 

.014623 

•65 

0.18098 

0.18617 

O .19578 

.008604 

.010718 

.014295 

.70 

0.13926 

0.14419 

0.15333 

.008182 

.010192 

•013595 

•75 

0.10226 

0.1068 

O. II519 

.00752 

.009367 

.012494 

.80 

O.O7033 

0.07431 

O.08368 

.00660 r 

.00S222 

.010967 

.85 

O.O4379 

0.04706 

O.O5309 

.005421 

.006752 

.009006 

.90 

0.02301 

0.02537 

O.02974 

.003919 

.004882 

.006512 

•95 

O.O0829 

0.009576 

O.OII945 

.002123 

.0026437 

.003527 

1.00 

0.00 

0.00 

0.00 

0.00 

0.00 

0.00 


either extremity of a three-span, rim-bearing drawbridge for 
a square crossing and for arms of equal length. The ratios, 
c y of either arm to the center span or panel, as shown 




















i8o 


SWING BRIDGES. 


immediately preceding Equation (3) of Article 35, are given 
in Table I, with the greatest, least, and mean values, taken 
from fifteen drawbridges as they have actually been designed. 
They cover, therefore, a range that will include nearly all 
practical cases of pin-connected structures. The columns of 
the table show the reactions R x and R 4 for a unit panel load 
placed successively at distances from the free end of the arm 
4 which vary by .054. These reactions due to unit panel 
loads are computed from Equations (5 a) and ( 6 a) of Article 



Fig. 1. 

35, with the values of c given at the heads of the columns of 
the table. As an example, if the unit load be placed at the 
distance 0.24 from the free end of the arm, the reaction R v 
will have the values 0.71355 and 0.72112 for the values of 
l -r- 4 = c — 8.822 and 4.395 respectively. The corresponding 
reactions, R A , will be .004401 and .007311. The actual reac¬ 
tions for any panel load will be found by simply multiplying 
those given in the table by the actual panel load in pounds 
or tons, as the case may be. 

It will be observed that the reactions vary very little 

















































TABLES AND DIAGRAMS. 


181 


% 

between the limits of range of the values of c. Hence, for 
values materially outside of those limits, the corresponding 
reactions may be assigned by the aid of Figures I and 2 with 
sufficient accuracy. Those figures exhibit the results given 
in Table I. Figure 1 gives the reactions R lf and Figure 2 the 



reactions R±. The three curves so nearly coincide in Figure I 
that only those for the two values of c — 8.822 and c — 4.395 
are shown. For practical working, that figure should be 
drawn to at least double the size shown. In Figure 1 the 
tabular values of the reactions R x have been taken full 
size, while those for R 4 , in Figure 2, have been multiplied 
















































































182 


SWING BRIDGES . 


by 10,000. The figures will enable either of the reactions to 
be read at a glance for any length of panel in any length of 
arm. They, as well as Table I., are, therefore, perfectly gen¬ 
eral for a wide range of the value of c. 

After the reactions are read from the table or the diagrams, 

moments for either the web or chord stresses can readily be 

* 

written ; and from these moments the stresses themselves 
will at once result in accordance with the methods of the 
preceding Articles. If the chords are parallel, moments will 
not be required for the web stresses, as the latter can be at 
once written from the shears. 

In many cases, particularly if the chords are parallel, the 
values given in Tables II. to VIII. will be found very conven¬ 
ient for seven- to nineteen-panel drawbridges, in which the 
center panel is equal in length to the others. They have 
been computed by Mr. Frank C. Osborn, consulting engineer, 
of Cleveland, Ohio, for his own practice, who has kindly given 
the author the privilege of using them in this connection. 
They express the shears and moments on the basis of each 
panel load being unity, and of each panel being one unit in 
length. Each actual shear will therefore be found by multi¬ 
plying each tabular shear by the actual panel load, and each 
actual moment will be found by multiplying each tabular 
moment by the product of the actual panel load by the actual 
panel length. These tables also show the greatest shears and 
moments at the various panel points. The reactions due to 
the various panel loads, by the aid of which the shears and 
moments are obtained, are computed from Equations (5^) 
and ( 6 a) of Article 35, except those which belong to the 
simple spans, which, of course, follow the law of the lever. 
If the chords are not parallel, the web stresses must be 
determined by the method of moments, as illustrated in the 
two preceding Articles. Even if the length of the center 
panel should differ to some extent from that of the others, 
Table I. and Figs. 1 and 2 show that the values in Tables II. 
to VIII. will not be sensibly changed. 


TABLES AND DIAGRAMS. 


183 


TABLE II. 


B C D E EG 



Note. —Shear in panel ab — reaction at a, and Shear cd = reaction at d. 


Loads at : 

Shear in Panel : 

Moment at : 

ab 

be 

cd 

b 

c 

d 

b and tr ... 

+ 0 568 
+ 0.210 

—0.432 
+ 0,210 

—0.432 

-0.790 

+ 0.568 
+ 0.210 

+ 0.136 
+ 0.420 

—0.296 

-0.370 

c and /. . 

Maximum .-j 

As a Simple Span. -j 

+ 0.778 

+ 0.210 
-0.432 

+ 0.333 
- 0.333 


+ 0.778 

+ 0.556 


— 1.222 

—0.666 

+1.000 

+ 1.000 

+ 1.000 

— 1.000 







TABLE III. 


D 


a 

A 


bed 

9 PANELS 


K T 


e f g h 7 

A A ALL EQUAL 



•Note. —Shear in panel ab = reaction at a, and Shear de — reaction at e. 


Loads at : 

Shear in Panel : 

Moment at : 

ab 

be 

cd 

de 

b 

c 

d 

e 

b and i . 

c and h . 

d and g . 

Maximum. - 

+ 0.665 
+ 0.364 
+ 0.131 

—0.335 

+0.364 

+0.131 

- 0.335 
—0.636 
+ 0.131 

- 0-335 
—0.636 
—0.869 

+ 0.665 
+ 0.364 
+ 0.131 

+ 0.330 
+ 0.728 
+ 0.262 

—0.005 
+ 0.092 

+ 0.393 

—0.340 

-0.544 

-0 476 

+1.160 

+0.495 

-0.335 

+0.750 

—0.250 

+ 0.131 
—0.97 1 

+ 0.250 
—0.750 

— 1.840 

— 1.500 

+1.160 

+ 1.320 

+ 0.485 
—0.005 

+ 1.500 

— 1.360 

( 

As a Simple Span .... -j 

+ 1.500 

+1.500 

+ 2.000 














































































































SWING BRIDGES. 


184 


TABLE IV. 



Note. —Shear in panel ab = reaction at a , and Shear ef — reaction at f. 


Loads at : 

Shear in Panel : 

Moment at : 

ab 

be 

cd 

de 

ef 

b 

c 

d 

e 

/ 

b and l . 

+ 0.726 
+ 0.471 
+ 0.252 
+ 0.089 

-c.274 
+ 0.471 
+ 0.252 
+ 0.089 

-0.274 
-0.529 
+ 0.252 
+ 0.089 

-0.274 
-0.529 
—0.748 
+ 0.089 

-0.274 
-0.529 
—0.748 
—0.911 

+ 0.726 
+ 0.471 
+0.252 
+ 0.089 

+ 0.452 
+ 0.942 
+ 0.505 
+ 0.178 

+ 0.178 
+ 0.412 

+ 0.757 

+ 0.268 

-0.095 
—0.117 
+ 0.009 
+ 0.357 

—0.369 

—0.646 

-0.738 

- 0.554 

c and k . 

d and i . 

e and h . 

Maximum .... -J 

As a Simple j 
Span. 

+ 1.538 

+ 0.812 
-0.274 

+ 1.200 

—0.200 

+ 0.341 

-0.803 

+ 0.600 
— 0.600 

+ 0.089 
-I- 55 I 

+ 0.200 

-1.200 


+ 1-538 

+ 2.077 

+ 1.615 

+ 0.366 
— 0.212 

+ 2.000 

-2.307 

— 2.462 

+ 2.000 

+ 2.000 

+ 3.000 

+ 3.000 

— 2.000 




1 







TABLE V. 

BCD E F G H 1 K L M N 



Note. —Shear in panel ab — reaction at a, and Shear fg = reaction at g. 

* 


Loads at : 

Shear in Panel: 

ab 

be 

cd 

de 

ef 

fs 

b and n . ... 

+0.768 

— 0.232 

—0.232 

—0.232 

— O. 232 

— 0.232 

c and m . 

+0.548 

+0.548 

—0.452 

—0.452 

—0.452 

—0.452 

d and l . 

+0.350 

+0.350 

+0.350 

—0.650 

—0.650 

—O.65O 

e and k . 

+0.185 

+0.185 

+0.185 

+0.185 

—0.815 

—0.815 

f and i . 

+0.065 

+0.065 

+0.065 

+0.065 

+ 0.065 

-°-935 

Maximum. 

+1.916 

+1.148 

+0.600 

+0.250 

+ 0.065 




—0.232 

—0.684 

- 1-334 

-2.149 

-3.084 

As a Simple Span. - 

+2.500 

+1.667 

+1.000 

+ 0.500 

+ 0.167 




—0.167 

—0.500 

— 1.000 

— 1.667 

— 2.500 

















































































































TABLES AND DIAGRAMS. 


I8 5 


Loads at : 

Moment at : 

b 

c 

d 

e 

f 

g 

b and n . 

+ 0.768 

+ 0.537 

+ 0.305 

+ 0.074 

—0.158 

— O.39O 

c and m . 

+ 0.548 

+1.097 

+ 0.645 

+ 0.193 

-0.259 

— O.7IO 

d and l . 

+ 0.350 

+ 0.700 

+ 1.050 

+ 0.400 

—0.250 

—0.900 

e and k .. 

+ 0.185 

+ 0.370 

+ 0.556 

+ 0.741 

-0.074 

—0.889 

/"and z. 

+ 0.065 

+ 0.129 

+ 0.194 

+ 0.259 

+ 0.323 

—0.612 

Maximum. 

+ 1.916 

+ 2.833 

+ 2.750 

+1.667 

+ 0.323 


i 





-0.741 

— 3-501 

As a Simple Span. 

+ 2.500 

+ 4.000 

+ 4.500 

+ 4.000 

+ 2.500 



TABLE VI. 



Note. —Shear in panel ab — reaction at a, and Shear gh = reaction at h. 


Loads at : 

Shear in Panel : 

ab 

be 

cd 

de 

ef 

fg 

gh 

b and p . 

+ 0.799 

—0.201 

— 0.201 

—0.201 

—0.201 

— 0.201 

— 0.201 

c and 0 . 

+ 0.606 

+0.606 

-0.394 

-0.394 

-0.394 

-0.394 

-0.394 

d and n . 

+ 0.427 

-(-0.427 

+ 0.427 

-0.573 

-o -573 

-0.573 

-0.573 

e and m . 

+ 0.270 

+ O.27O 

+0.270 

+0.270 

-0.730 

—0.730 

—0.730 

f and l . 

+ 0.142 

+0.142 

+0.142 

+0.142 

+ 0.142 

—0.858 

—0.858 

^•and k . 

+ 0.049 

+0.049 

+ 0.049 

+0.049 

+ 0.049 

+0.049 

-0.951 

Maximum . . -j 

+ 2.293 

+ 1.494 

+ o .828 

+0.461 

+ 0.191 

+ 0.049 




— 0.201 

-o -595 

—1.168 

— 1.898 

-2.756 

- 3 - 7°7 

As a Simple^ 

+ 3.000 

+ 2.143 

+1.429 

+0.857 

+ 0.429 

+0.143 


Span. t 


-0.143 

— O.429 

-0.857 

-1.429 

—2.143 

— 3.OOO 


Moment at : 


Loads at : 

b 

c 

d 

e 

/ 

g 

h 

b and p . 

+ 0.799 

+ 0.600 

+ 0.398 

+ 0.198 

—0.003 

— 0.203 

— O.4O4 

c and 0 . 

+ 0.606 

+1.212 

+ 0.819 

+ 0.425 

+ 0.031 

— 0.363 

- 0.757 

d and n . 

+ 0.427 

+ 0.855 

+1.282 

+ 0.709 

+ 0.137 

-0.436 

— I .009 

^and m . 

+ 0.270 

+ 0.540 

+ 0.811 

+1.081 

+ 0.351 

- 0.379 

— I . 109 

f and / . 

+ 0.142 

+ 0.283 

+ 0.425 

+ 0.566 

+ 0.708 

—0.150 

— 1.009 

^-and k . 

+ 0.049 

+ 0.098 

+ 0.148 

+ 0.197 

+ 0.246 

+ 0.295 

—0.656 

Maximum. j 

+ 2.293 

+ 3-588 

+ 3-883 

+ 3.176 

+ 1.470 
— O.OO3 

+ 0.295 
— I - 53 1 

- 4.944 

As a Simple Span.. 

+ 3.000 

+ 5.000 

+ 6.000 

+ 6.000 

+ 5.000 

+ 3.000 


































































































































SWING BRIDGES. 


186 


TABLE VII. 



Note.—S hear in panel ab = reaction at a , and Shear hi — reaction at i. 


Shear in Panel: 


Loads at : 

ab 

be 

cd 

de 

ef 

fg 

gh 

hi 

b and r . 

+ 0.823 

-0.177 

-0.177 

-0.177 

-0.177 

—0.177 

-0.177 

-0.177 

c and q . 

+ 0.651 

+0.651 

- 0.349 

- 0.349 

- 0-349 

- 0-349 

- 0.349 

- 0-349 

d and p . 

+ 0.489 

+0.489 

+ 0.489 

—0.511 

—0.511 

—0.511 

—0.511 

—0.511 

e and o . 

+ 0.342 

+0.342 

+ 0.342 

+ 0.342 

—0.658 

—0.658 

—0.658 

—0.658 

/ and n . 

+ 0.215 

+0.215 

+ 0.215 

+ 0.215 

+ 0.215 

-0.785 

-0.785 

-0.785 

^•and m . 

+ 0.112 

+0.112 

+ 0.112 

+ 0.112 

+ 0.112 

+ 0.112 

-0.888 

-0.888 

h and 1 . . 

+ 0.039 

+0.039 

+ 0.039 

+ 0.039 

+ 0.039 

+ 0.039 

+ 0.039 

—0.961 

Ma virrmm . < 

+ 2.671 

4 * 1.848 

+I.IQ 7 

+ 0.708 

+ 0.366 

+ 0.151 

+ 0.039 




1 

0 

M 

VI 

VJ 

—0.526 

-1-037 

-1.695 

0 

zo 

Tf- 

Ci 

1 

- 3-368 

— 4 - 3 2 9 

Act ^imnlp f^nan X 

+ 3 • 5°o 

+ 2.625 

+ 1.875 

+ 1.250 

+ 0.750 

+0.375 

+ 0.125 




—0.125 

-0.375 

-0.750 

— 1.250 

-1-875 

—2.625 

— 3.500 


Moment at: 


LOADS AT : 

b 

c 

d 

e 

f 

g 

h 

i 

b and r . 

+ 0.823 

+ 0.646 

+ 0.470 

+0.293 

+ 0.116 

—0.061 

—0.238 

—°- 4 I 5 

c and q . 

+ 0.651 

+ 1.303 

+ 0.954 

+0.605 

+ 0.257 

—0.092 

-0.441 

—0.789 

d and p . 

+ 0.489 

+ 0.979 

+ 1.468 

+0.957 

+ 0.447 

—0.064 

- 0.575 

— 1.086 

e and 0 . 

+ 0.342 

+ 0.684 

+ 1.026 

+1.369 

+ 0.711 

+ 0.053 

—0.605 

— 1.263 

/ and . . 

+ 0.215 

+ 0.430 

+ 0.644 

+0.859 

+ 1.074 

+ 0.289 

-0.497 

— 1.282 

^and in . 

+0.112 

+ 0.224 

+ 0.336 

+0.448 

+ 0-559 

+ 0.671 

—0.217 

— 1.105 

h and l .. 

+ 0.039 

+ 0.077 

+ 0.116 

+0.155 

+ 0.193 

+ 0.232 

+ 0.271 

—0.691 

Maximum. -j 

+ 2.671 

+ 4-343 

+ 5-014 

+ 4.686 

+ 3-357 

+ 1.245 
—0.217 

+ 0.271 
- 2-573 

—6.631 

As a Simple Span. 

+ 3 - 5 °° 

+ 6.000 

+ 7 - 5 °° 

+ 8.000 

+ 7.500 

+ 6.000 

+ 3.500 









































































































TABLES AND DIAGRAMS. 



TABLE VIII. 



Note,— Shear in panel ab — reaction at a , and Shear ik — reaction at k. 


Shear in Panel: 


Loads at : 



ab 

be 

cd 

de 

ef 

fg 

g h 

hi 

ik 

b and t . 

* 4 " o. 842 

—0.158 

—0.158 

1 

0 

W 

Cn 

00 

—0.158 

—0.158 

—0.158 

00 

10 

H 

6 

1 

00 

10 

H 

6 

1 

c and ^. 

+0.687 

+ 0.687 

-0.313 

—0.313 

-0.313 

—0.313 

-o. 3>3 

-0.313 

-0.313 

d and r . 

+0.540 

+ 0.540 

+0.540 

—0.460 

—0.460 

—0.460 

—0.460 

—0.460 

—0.460 

e and q . 

+0.403 

+ 0.403 

+ 0.403 

+ 0.403 

-0.597 

— 0.597 

- 0.597 

- 0.597 

-0.597 

/and ‘p . 

+0.280 

+ 0.280 

+ 0.280 

+ 0.280 

+ 0.280 

— 0.720 

— 0.720 

-O.72O 

—0.720 

g and o . 

+0.175 

+0.175 

+0.175 

+0.175 

+ 0.175 

+ 0.175 

—0.825 

—0.825 

—0.825 

h and n . 

+0.091 

+ 0.091 

+ 0.091 

+ 0.091 

+ 0.091 

+ 0.091 

+ 0.091 

— O.9O9 

—0.909 

i and m . 

+0.031 

+ 0.031 

+ 0.031 

+ 0.031 

+ 0.031 

+ O.OjI 

■ +0.031 

+ 0.031 

—0.969 

Maximum... j 

+3.049 

+ 2.207 

+ 1 520 

+ 0.980 

+ 0.577 

+ 0.297 

+ 0.122 

+ 0.031 



. 

—0.158 

-0.471 

-0.931 

— 1.528 

— 2.248 

— 3-°73 

-3.982 

—4.95 1 

As a Simple j 

+4.000 

+ 3.1:1 

+ 2.333 

+1.667 

+1.111 

+ 0.667 

+ 0.333 

+ 0. hi 


Span. 1 

. 

—0. Ill 

-°-333 

—0.667 

— 1.111 

— 1.667 

- 2.333 

— 3.hi 

—4.000 


Loads at 




Moment at : 



• 

b 

c 

d 

e 

/ 

g 

h 

i 

k 

b and t . 

+ 0.842 

+ 0.684 

+ 0.526 

+ 0.368 

+ 0.210 

+ 0.052 

—0.106 

—0.264 

-0.423 

c and s . 

+ 0.687 

+ I -374 

+1.061 

+ 0.748 

+ 0.435 

+ 0.122 

— O. I9I 

-0.504 

—0.814 

d and r . 

+ 0.540 

+1■080 

+1.620 

+1.160 

+ 0.700 

+ 0.240 

— 0.220 

—0.680 

— I . 142 

e and q . 

+ 0.403 

+ 0.806 

+1.209 

+1.612 

+ 1.015 

+ 0.418 

-0.179 

—0.776 

— 1 -375 

/and p . 

+ 0.280 

+ 0.560 

+ 0.840 

+1.120 

+1.400 

+ 0.680 

—O.O4O 

—0.760 

— 1.482 

g and 0 . 

+ 0.175 

+ 0.350 

+ 0.525 

+ 0.700 

+ 0.875 

+ 1.050 

+0.225 

—0.600 

— 1 428 

h and n . 

+ 0.091 

+ 0.182 

+ 0.273 

+ 0.364 

+ 0.455 

+ 0 546 

+0.637 

— 0.272 

— 1.184 

i and m . 

+ 0.031 

+ 0.062 

+ 0.093 

+ 0.124 

+ 0.155 

+ 0.186 

+0.217 

+ 0.248 

-0.719 

Maximum.. . -j 

+ 3-049 

+ 5.098 

+ 6.147 

+ 6.196 

+ 5-245 

+ 3.294 

+ 1.079 
—0.736 

+ 7.000 

+ 0.248 
-3.856 

+ 4.000 

-8.567 

As a Simple | 
Span. j 

+ 4.000 

+ 7.000 

+ 9.000 

+10.000 

+10.000 

+ 9.000 


The drums of rim-bearing turntables should be of sufficient 
depth to prevent upward deflection from materially disturb¬ 
ing a uniform distribution of load over the rollers. The 













































































































SWING BRIDGES. 


1 8 7 a 

upward pressure of the latter constitutes an upward loading 
on the lower flange of the drum, and the points on the 
upper flange of the latter, at which the truss load is applied, 
form the supporting points of the continuous drum girder. 
The loading on the lower drum flange should be as nearly 
uniform as possible, and in order to secure that result, the 
drum depth should be as great as possible. It is also neces¬ 
sary that the truss load should be carried to the upper drum 
flange at as many equidistant points as may be found prac¬ 
ticable. In long and heavy draw spans it will be necessary 
to carry the truss loads to the drum through a combination 
of transverse and longitudinal girders, in order to secure the 
requisite number of points of application on the upper flange. 
A calculation of the girder strength of the drum can be made 
by assuming that its segments are beams with a span length 
equal to the distance between points of application of the 
truss loads on the upper flange, and that they are loaded 
with the uniform roller pressure. Although the drum is con¬ 
tinuous, these beams should be considered non-continuous, 
for they are not straight, and a failure to secure the assumed 
uniformity of loading may essentially destroy the advantages 
of continuity. The results of all such computations must, 
however, be strongly tempered with those of experience. 
The drum section must be such as to avoid any appreciable 
deflection ; its depth should never be less than one-third the 
distance between adjacent points of support on the upper 
flange, and one-half is better practice. If the total truss 
load does not require nearly all the rollers which the circum¬ 
ference of the drum affords, the rule may be proportionately 
modified, but not otherwise. 

The Thames River bridge, carrying about 2,400,000 pounds 
dead load on the rollers, has eight equidistant points of sup¬ 
port on its drum, with 32 feet diameter and 5 feet depth. 

The 500 feet single-track Arthur Kill draw span has about 
1,450,000 pounds of dead load resting at eight equidistant 
points on a drum 27! feet in diameter and 3! feet deep. 

The Central bridge across the Harlem River at New York 
has two concentric drums 44 feet and 36 feet in diameter 


DRUMS AND ENGINES. 


i Syb 

with sixteen points of support on each. The total truss 
weight is about 3,800,000 pounds, and the depth of the drum 
is 5 feet. The total truss weight of the Kingsbridge Road 
bridge across the Harlem Ship Canal at New York is about 
1,500,000 pounds, and it is carried at twelve equidistant points 
of support on a drum 39 feet in diameter and 5 feet deep. 

It is usually very easy to secure eight equidistant points of 
support on the drum of an ordinary single-track draw span, 
with a depth of drum of at least 30 to 36 inches, and such 
an arrangement should be required. The resulting distri¬ 
bution of load on the rollers will be found very satisfactory. 

The power required to be exerted by an engine to turn a 
drawbridge is expended in three parts. One portion is used, 
at the beginning of the operation of opening or closing, in 
developing the maximum velocity possible, and is stored for 
a short time as the actual energy of the structure in motion ; 
it subsequently performs work against some brake arrange¬ 
ment by which the bridge is brought to rest. A second por¬ 
tion is used in performing work against the entire frictional 
resistance of the moving parts of the structure and machin¬ 
ery ; while the third and last portion is required to overcome 
the wind resistance when the wind blows against one arm 
with a total pressure which is not balanced by that against the 
other. The rolling friction, or, rather, the entire friction, of a 
drawbridge in motion, has been determined by Mr. Theodore 
Cooper for the Second Avenue double-track railroad bridge 
over the Harlem River at New York; and for the Thames 
River double-track railroad bridge at New London, Conn., by 
Messrs. Boiler and Schumacher; and the results of these in¬ 
vestigations can be found in the Transactions of the Ameri¬ 
can Society of Civil Engineers, for December, 1891. The 
total moving weight of the Second Avenue bridge was 880,- 
000 pounds, and that of the Thames River bridge 2,400,000 
pounds. The length of the latter is 500 feet, and the diame¬ 
ter of the drum 32 feet. Mr. Cooper found the coefficient 
of frictional resistance for the Second Avenue bridge to be 
.0038— i.e. } a force of 3.8 pounds per 1,000 pounds of total 
weight moved would have to be applied tangentially at the 


SWING BRIDGES. 


18 yc 

center line of the track (or to the drum) in order to overcome 
the total friction. 

Messrs. Boiler and Schumacher found a coefficient of 
about .004 for the Thames River bridge, or 4 pounds per 
1,000 pounds moved. The greater part of the “total fric¬ 
tion ” is the rolling friction at the drum. These two instances 
are the most valuable rim-bearing drawbridge investigations 
of the kind ever made in this country, and as the workman¬ 
ship, fitting of track, etc., were of an unusually excellent 
character, it is probably advisable to take the coefficient of 
friction for ordinary draw spans at .01, or 10 pounds per i,oco 
pounds of weight moved. 

In the Transactions of the American Society of Civil 
Engineers for 1874, Mr. C. Shaler Smith gave the results of 
a number of less complete but very interesting tests of rim¬ 
bearing draw spans in about the ordinary conditions of work¬ 
manship and running order. He found the total friction to 
vary from 4 to 8 pounds per 1,000 pounds of weight moved. 

The power required to give the desired velocity of rotation 
to a drawbridge will depend upon the time allowed for open¬ 
ing or closing. Draws operated by power are usually 
opened, or closed, in one to three minutes. Small draws 
operated by hand will consume three to eight minutes. 

If drawbridges are operated against an unbalanced wind 
pressure, the necessary power increases very rapidly. Seven- 
eighths, or even nine-tenths, of the total capacity of a well- 
proportioned drawbridge engine maybe exerted against wind 
pressure when the structure is moved in a moderately high 
wind. A comparatively small amount of power is required 
to overcome the friction. 

In the 500 feet double-track Thames River bridge, with 
moving parts weighing about 2,400,000 pounds, not more 
than 5 or 6 horsepower, at most, was expended in develop¬ 
ing the acceleration and overcoming the total friction. An 
unbalanced wind pressure of 5 pounds per square foot on one 
arm would have required only a little less than 30 horsepower 
to turn the draw against it, or double that amount for a 
io-pound wind. 


DRUMS AND ENGINES. 


1 8? d 


The computation of the work required to turn a draw¬ 
bridge will require its moment of inertia to be taken about a 
vertical axis through the center of the drum. It will be suf¬ 
ficient for this purpose to consider the trusses, lateral brac¬ 
ing, floor system, and track as a homogeneous prism with 
length / and width w. This portion of the weight is, for all 
practical purposes, five-sixths the total moving weight W for 
single-track railroad spans, and seven-eighths the same total 
weight for double-track spans. Hence the moment of inertia 
I' of this portion of the weight will be—if g is the approxi¬ 
mate constant, 32.2, for gravity—for single track spans : 


r _ | w(w* + / 3 ) = 5 w(w 2 + r) 

12 g ~ 72g 



Or for double-track spans: 

_ ; W (w 2 + / 2 ) 

9% 



The moment of inertia of the drum, rollers, etc., can be 
considered concentrated at the distance R = radius of the 
drum, from the axis. Hence the moment of inertia, /", of 
this portion of the weight will be, for single-track spans: 


WR 2 



Or, for double-track spans : 


I 


rt 


WR 2 



Hence the total moment of inertia will be : 

/=/' + /".( 5 ). 

The nominal horsepowers of engines fitted to a number of 
railroad and heavy city draw spans which have proved to 









SWING BRIDGES. 


\Zje 

be very satisfactorily operated are given in the tabulation 
below : 


Bridge. 

Weight of 
moving parts. 

Engines. 

500 feet double track railway. 

LBS. 

2,400.000 

H.P. 

40 

400 feet city bridge. 

4,200,000 

50 

270 feet city bridge. 

1,800,000 

40 

400 feet double track railway. 

2,000,000 

35 

300 feet double track railway. 

1,250,000 

20 

362 feet single track railway. 

684,000 

20 

217 feet double track railway. 

600,000 

20 


The tendency has been toward an increase of engine 
power, in consequence of some of the earlier and smaller 
engines having shown insufficient capacity in winds and 
other contingencies of drawbridge operation. 

The preceding considerations apply to rim-bearing turn¬ 
tables only, which are now exclusively used for all draw 
spans over a length sufficiently great to require the structure 
to be of the “through ” type. 

The center- or pin-bearing type, in which the entire 
moving weight of the structure is carried on a center pin 
or pivot, is used for short spans only. This center pin 
or pivot may vary from eight to twelve or more inches 
in diameter, so that the pressure per square inch of bear¬ 
ing surface will not take greater values than two thousand 
pounds to twenty-five hundred pounds. The center pin, 
of wrought iron, or, preferably, steel, frequently rests upon 
a flat disk of phosphor-bronze in order to reduce friction 
and w T ear. The bearing faces of the pin and disks should 
be channeled or grooved to allow entrance for a heavy 
lubricant. 

In the Transactions of the American Society of Civil En¬ 
gineers for 1874, Mr. C. Shaler Smith gave, as the result 
of a number of tests of center-bearing draw spans, the fric¬ 
tional resistance, if exerted at the circumference of the center 
pin , at r tfo of the total weight turned. If W is that weight 

















DRUMS AND ENGINES. 1 87/ 

in pounds, and F the force of friction supposed exerted in 
the circumference of the pin, then : 

F= .09 W .(6). 


If, again, t is time in minutes required by one man to open 
the draw, or close it, and if d is the diameter of the center 
pin in feet, he gave : 


t — .09 W 


nd 


4 x 10,000 



The time given by Equation (7) will usually be insufficient 
for the requirements of one man, even if no other considera¬ 
tion than that of friction be involved. 

In the contingency of an unbalanced wind pressure on one 
arm of the draw, one man may not be able to put the bridge 
in motion, nor to hold it against the wind. Hence none but 
the smallest draws in unimportant situations are made de¬ 
pendent on one man power. 

On account of the intermittent character of the operations 
of the motive power of a draw-bridge, gas or hot air en¬ 
gines are admirably adapted to that purpose. Their increased 
economy and essential cessation of expenses when the bridge 
is not being turned meet the requirements of the situation in 
a very satisfactory manner. As far as they have been tried 
they leave little to be desired. 


Art. 38 .—Ends Simply Resting on Supports—One Support at Centre— 

Example. 

The general principles fundamentally involved in this case 
are not different from those of the two preceding ones, except 
in the number of points of support at the first pier. All the 
fixed load of the bridge is carried to the central point of sup- 




188 


SWING BRIDGES. 


port, whether the bridge is open or closed; the end supports 
furnish reactions for the moving load only. 

The truss to be taken as an example is the one shown in 
the accompanying figure, in which the arms are of equal 
length. 

The general formulae to be used for the reactions at A, B, 
and C, and for the bending moment at the centre, are equa¬ 
tions (u), (12), (13), and (10) respectively of Art. 35. These 
equations may be written as follows, remembering that l x = 
4 = /, and M 2 = M : 

M=-I-jSp(P-g‘)s+2P(l 2 -z ! )zl. . (1). 


= hp{l-z)+M |. (2). 

k + 2 Ps - iM j..(3). 

R 3 =l^ 2 P(l- s )+Ml . (4). 


It is to be remembered that z is measured from A or U, 
according as the left or right arm is considered. 



The following are the data to be used : 

Total length — AC — 2 1— 2AB = 2BC — 144 feet. 
Uniform depth = dD — bB — 16 feet. 

Panel length = AD = DE — etc. = 13 feet. 

BH = BH’ = 7 feet. 


















ENDS SIMPLY RESTING ON SUPPORTS. 


189 

Total fixed weight per foot = 1,200 pounds (nearly). 

Upper chord panel fixed weight = W = 2.73 tons. 

Lower “ “ “ “ = W — 5.00 “ 

Uniform panel moving load = w — 19.50 “ 

The moving load traverses the lower chord, and the weight 
of the floor system is taken at nearly 350 pounds per foot. 

On account of the extra weight of the locking apparatus, 
the fixed weight at A, or C, will be taken at 3 tons, and will 
be denoted by w v 

As is clear from the figure, all inclined web members, ex¬ 
cept the end posts, are for tension only, while the verticals 
are compression members. 

As the ends A and C are neither latched down nor lifted 
up, either arm is a single truss simply supported at each end, 
for all moving loads which rest upon it, so long as there are no 
moving loads on the other arm. 

For exactly the same reasons, therefore, as those given in 
the preceding Article, any counter , as dE, will sustain its 
greatest stress when the moving load extends from its foot to 
the centre, if no other moving load rests on the bridge. 

It must still be borne in mind that in connection with any 
counter stress, the stress in the vertical which cuts its upper 
extremity is to be found, for such a one may be the greatest 
stress in the vertical. 

Again, resuipe the general expression for the shear in any 
web member: 

s = S — n( W + W) — nw ; 

in which S is the shear at one extremity of the arm, and n 
and n the numbers of fixed and moving panel weights re¬ 
spectively between the same end of the arm and the web 
member in question. In considering the main web members, 
S will be taken adjacent to the centre, and, in the present 
example, at an indefinitely short distance from B in the arm 
AB. 

For a given condition of loading in AB, it is evident that 
the smaller is R\ the greater will be S. But Eq. (1) shows 


SWING BRIDGES. 


I90 

l 

that M is always negative. Hence so long as 2 P(l — s) re¬ 
mains the same, Eq. (2) shows that decreases as M in¬ 
creases (numerically). 

Again, Eq. (1) shows that M will have its greatest numerical 

2 

value, other things remaining the same, when 2 P (l 2 — P)z 

has the greatest value possible; i. e., when the moving load 

covers the whole of the arm BC. With a given value, there- 
1 

fore, for ^P(l— z), will be the least possible when the 
moving load covers the whole of the other arm, or the whole 
of BC; consequently 5 will be the greatest under the same 
conditions. Now having found under what circumstances 5 
is the greatest, precisely the same reasoning used in the pre¬ 
ceding Articles shows that ^ will be the greatest, under the 
same circumstances, when n is zero. 

A ny inclined main web member, then, will sustain its greatest 
tensile stress when the moving load extends from its foot to the 
free extremity of the arm in which it is found, and covers at the 
same time the whole of the other arm. 

A few main web stresses in all trusses of this case are a 
little singular in character, but are no exceptions to this rule. 
Those for the example taken will be noticed in the proper 
place. 

Any vertical web member, unless acting as a counter, will 
sustain its greatest compression in connection with the great¬ 
est tension in the inclined main web member which cuts its 
upper extremity. 

In seeking the greatest main web stresses, it may happen 
that the reaction R x becomes zero; this, however, changes 
nothing in the method. 

Since either arm is a simple truss for all moving loads rest¬ 
ing on it (supposing none on the other), every such load tends 
to cause the same kind of stress throughout the same chord. 
Consequently, as in the previous Article, the greatest tension in 
the lower chord, and compression in the tipper, will exist when the 
moving load covers one arm only. These stresses will be found 
in that portion of the arm adjacent to its free extremity. 

The greatest chord stresses of the same kind as those 


ENDS SIMPLY RESTING ON SUPPORTS. 


l 9 l 

caused by the fixed load, can be found with the least labor 
by first determining all the stresses due to the fixed load alone , 
and tabulating them. 

The stresses caused by the moving load alone are then to 
be determined by the aid of the following considerations. 

Let that moment be considered negative which causes 
tension in the upper chord and compression in the lower. 
All moments, then, caused by the fixed or moving loads are 
negative, and all those produced by the upward reactions R 1 
are positive. Now the compression in any lower chord panel 
may be found by taking moments (such will be negative if 
compression exists) about the panel point vertically over that 
extremity nearest the centre. The general expression for 
such compression will be: 

R x np — nwt 
d : 

in which n is the number of the panel from the free extremity 
of the arm, 71 the number of the moving panel loads on the 
arm, t the distance of the centre of gravity of the moving 
loads 71 w from the origin of moments, / the panel length, and 
d the depth of the truss. 

The numerator of this expression must be negative, and it 
is desired to find what value of 11 will give it its greatest 
negative value. 

Now since every panel moving load on the arm AB 
increases R x (as a positive quantity), it appears from the 
figure that n must not be greater than (n — 1), and, farther, it 
must belong to loads between the panel considered and the 
free extremity of the arm. Since, however, t varies with ri 
the above expression may have its greatest negative value 
when 71 is less than (n — 1). 

These considerations are independent of the general charac¬ 
ter of Rg, it has already been seen, however, that, with a 
given loading on AB, R t will be the least when the moving 
load covers the whole of BC also. Hence in order to find the 
greatest tension in the upper chord and compression in the 



192 


SWING BRIDGES. 


lower due to the moving load, the method of procedure is as 
follows: 

Throughout the whole operation the moving load is to 
entirely cover one arm, as BC. The moving load is then to 
cover the other arm from the free extremity to any panel 
point, and the stresses in the panels situated between 
the end of the train and the centre are to be computed. 
This operation is to be repeated for every panel in that arm 
not wholly covered by the moving load. From these results 
the greatest stresses may be selected and then added to the 
fixed load stresses. 

The character of these operations and the reasons for them 
will be much more evident after the example is treated. 

The following values will be needed, and depend only on 
the data already given. 


*=13 

. . 1 — z — 59 . 

. (/ 2 — P)z = 65,195.00 

II 

to 

Os 

• 

VO 

II 

1 

• 

• 

. (l 2 — P)z — 117,208.00 

* = 39 

. . l-z= 33 . 

. (/ 2 — P)z = 142,857.00 

* — 5 2 

. . 1 — Z — 20 . 

. (I 2 — P)z — 128,960.00 

*= 65 

• 

IN. 

II 

1 

• 

• 

• ( /2 - £) z — 62,335.00 


Angle AdD 

= a. 


“ HbB 

= /?. 


tan a = 0.8125. 

sec a — 1.29. 


tan ft = 0.4375. 

sec ft = 1.09. 


1 _ 1 

4 l 2 20736 " 

= 0.00004823 

P — 19.5 tons = w; 

p 

7 = °- 271 • 


The following values of M are found by substituting the 
proper numerical values in Eq. (1). The moving load is taken 
to cover the whole of BC and so much of AC as is indicated 
by the values of z. 







ENDS SIMPLY RESTING ON SUPPORTS. 


193 


In arm AB ; z — 

u u u . „ _ 

, 4 — 



M— - 547.00 
M — — 657.00 


(S 


u 


; z 


( *3 
1 26 

39 


I 


u u 


u 


; z 


13 

26 

39 

52 


(( u 


a 


; z 




13 

26 

39 

52 

65 


M — — 792.00 


M — — 913.00 


M — — 972.00 


The following values may now be written. Those of R x 
are found by simply substituting the proper numerical quanti¬ 
ties in Eq. (2) ; the moving load, as the values of M show, is 
taken to cover the whole of BC. 


z- 13 


z — 



P 1 

. — 2 (l — z) = 15.99 tons • • Bi = 8.39 tons. 
“ = 28.46 “ . . R x = 19.33 “ 


( 13 

Z = -l 26 . . “ = 3740 “ . . R x 26.4O 

( 39 



'13 

26 

39 

152 




13 

26 

39 

52 

165 

13 


= 42.82 “ . . R x = 30.14 


= 44.72 “ . . R x = 31.22 








l 94 


SWING BRIDGES. 


The stresses in the counters will first be sought, i.e ., those 
in the arm AB. 

As a trial let the moving load cover the points F, G, and 
H, and, as before, let B 1 be the general expression for the re¬ 
action at A. Hence: 

x _l 7 

R x — 3 x 19.5 x - D ^ - = 16.25 tons. 

Since 3 + 2 x 7.73 > R t , 'EP = o at the panel point E , and 
no counter is needed between e and F. 


Moving load over EH. 


R \ 4 x 19.5 x 


26.5 

72 


= 28.71 tons. 


As 3 + 2 x 7.73 + 19.5 > R x , 2 P= o at E , and dE is the 
first and only counter needed. 

The vertical component of the stress in dE is 

s = R x — 3 -f 7.73 — 17.98 tons. 

Hence, (dE) = 17.98 x sec a — 4- 23.19 tons. 

% 

The greatest compression in the end post Ad, and tension 
in the vertical d'D will exist when the moving load covers the 
whole of the arm AB, the other carrying none, and with such 
loading: 

2 -3 

R x = 5 x 19.5 x — = 44.69 tons. 


Hence, (Ad) — — (R x — 3) x sec a — — 53.78 tons. 
At the same time: 


(d'D) — + (19.5 + 5.00) = +24.5 tons. 

The stresses in the main web members are next to be de¬ 
termined. 

Those in Ad and d'D will occur under circumstances to be 
indicated hereafter. 




ENDS SIMPLY RESTING ON SUPPORTS. 


195 

The following operations are in accordance with the princi¬ 
ples already shown. 

Moving load on BC and at D. 

R\ = 8.39 tons; hence, for the shear in De : 

^ = — R x + (3 + 7.73 + 19.5) = 21.84 tons. 

Hence, (ZV) = s x sec a = + 28.17 tons. 

Also, (eE) — — (s + 2.73) = — 24.57 tons. 

Moving load on BC and DE. 

R\ — 19.33 tons ; hence, for the shear in Ef: 

s ■- - R x 4* ( 3 + 2 x 7.73 + 2 x 19.5) = 38.13 tons. 

Hence, (Ef) = s x sec ol — + 49.19 tons. 

Also, (/A) = — (s + 2.73) = — 40.86 tons. 

Moving load on BC aiid DF. 

R t = 26.40 tons; hence, for the shear in Eg: 

s=—R i+(3 + 3 x 7.73 + 3 x 19.5) = 58.29 tons. 

Hence, (A*r) = s x sec a = + 75.19 tons. 

Also, ( gG ) = — (s + 2.73) = — 61.02 tons. 

Moving load on BC and DG. 

R l = 30.14 tons; hence, for the shear in Gh : 

s = —R l + ( 3+4X 773 + 4 x 19 - 5 ) = 8 l 7 § tons. 

Hence, (Gli) = s x sec a = 4 - 105.5 tons. 

Also, ( 777 ) = — (j + 2.73) = - 84.51 tons. 

Moving load over B C and A B. 

R x = 31.22 tons ; hence, for the shear in Hb : 

s = - R t + (3 + 5 x 7.73 + 5 x 19.5) = 107.93 tons. 


196 


SWING BRIDGES. 


Hence, (Hb) — s x sec /3 = + 117.64 tons. 

The same panel weights have been taken for H and H' as 
for D , E, E, etc., though, strictly speaking, they would be a 
little smaller. At b, however, the fixed weight will be taken 
as 2.73 x 7 -r- 13 =1.47 tons. 

Hence, (bB) = — (2 x 107.93 + 1.47) = — 217.33 tons. 

Thus the web stresses, with the exceptions noticed, are 
completed. 

It has been shown that the greatest compression in the up¬ 
per chord and tension in the lower will exist when the moving 
load covers the whole of one arm, as AB, for which condition 
of loading, as has already been seen: 

E-l = 44.69 tons. 

Now, 7 ^ — (3 + 7.73 + 19.5) = 14.46 tons is that part of the 
total panel load (fixed and moving) at E, which may be con¬ 
sidered as passing directly to A ; while (19.5 + 7-73) — 14.46 
= 12.77 t° ns is the remainder, which maybe taken as passing 
directly to B . 

The following values will now be needed: 

14.46 x tan a = 11.75 tons. 

12.77 x tana = 10.37 “ 

(7.73 + 19.5) x tana — 22.12 “ 

The chord stresses then follow: 

{AD) — ( DE ) = ( 7 ?i — 3) x tan a — 4- 33.87 tons. 

(de) — (ef) = — (2 x 14.46 + 27.23) x tan a = — 45.62 tons. 
(fg) = (ef) + 12.77 x tana- — 35.25 tons. 

(EF) - - ( fg ) = + 35.25 tons. 

(gE) - (fg) + 27.23 x tan a + 12.77 x tan a = — 2.76 tons. 

(FG) — — (g/i) = + 2.76 tons. 


ENDS SIMPLY RESTING ON SUPPORTS. 


19 7 

(Jib) will evidently be tension, and ( GH ) compression; no 
other stresses, therefore, are needed. 

The following stresses, by moments, serve as checks: 

( de ) = (,/) = - (*■ ~ 3 ) * 2 7- 2 3 * *3 = _ 45 . 62 tons . 

{ gh ) = - ■ <*« ~ 3) * J * . 4 x 27.23 x 19.5 = _ 275 tons . 

The chord stresses due to the fixed load alone are the fol¬ 
lowing: 

(AD) — — (de) — — 3 x tan a = — 2.44 tons. 

(DE) = — (ef) — (AD) — (3 + 7.73) x tana — — 11.16 “ 

(EF) — — (fg) — (DE) — (3 + 2 x 7.73) x tan a — — 26.16 “ 

( FG ) = - (^) = (£A) - (3 + 3 x 7-73) x « = — 4744 “ 

(( 777 ) = — (bJi) — (EG) — (3 + 4 x 7.73) x tan a — — 75.00 “ 

( 7 //) = (( 7 //) - (3 + 5 x 7.73) x tan (3 = - 93 * 22 “ 

As a check : • 


(BH) = - 


3 X 72 + 5 X 7.73 X 33 

16 


93.21 tons. 


The tension in the upper chord and compression in the 
lower, due to the moving load only, still remain to be found. 

* 

Moving load over A B and B C. 

*i = 3 i .22 tons. Moments about b give: 

(BH)' — ^ x 7 2 ~ 5 x * 9-5 x 33 _ _ _ 5 0 5 tons. 

16 


Moving load over BC and A G. 
R t — 30.14 tons. Moments about h give : 


(GH)'— - (hby = 


Ri x 65 - 4 x 19.5 x 32.5 
16 


35.99 tons. 







198 


SWING BRIDGES. 


Moving load over BC and AF. 
R-l = 26.40 tons. Moments about g give : 



R t x 52 — 3 x 19.5 x 26 
16 


— 9.26 tons. 


Moving load over BC and AE. 
R x — 19.33 tons. Moments about /give: 


(FFy = - (fgy= 


Ri X 39 - 2 X 19.5 


16 


x 19.5 

-= — 0.4 tons. 


Moving load over BC and at D. 
R x — 8.39 tons. Moments about e give: 


{ DE)' = - (,/)' = * 26 


— 2.21 tons. 


(EF)' = —(fg)' = —2.21 — (19.5 — 8.39) x tan ol— — 11.24 tons. 
( FG )' = — (g/i)' = — 11.24 — 9.03 = — 20.27 tons. 

( 677 )' = — (lib)' = — 20.27 — 9.03 = — 29.30 tons. 

Other chord stresses, with the different conditions of load¬ 
ing taken, are not indicated, as they were found to be less, for 
the same panels, than those that are given. They might be 
needed, however, in some cases. 

A very important result occurs, which has not before been 
noticed, when the moving load covers AO and one panel load 
rests at A. 

In such a case Eq. (1) gives : 

M = - i (P - s 2 ) z = - 486.00. 

And Eq. (2): 

-r, M 

Ri = -j- — — 6.75 tons. 

Under the circumstances just named, therefore, the condi- 








ENDS SIMPLY RESTING ON SUPPORTS. 


l 99 


tion of things at A is equivalent to hanging a weight of 6.75 
tons at that point, as the end of the overhanging arm AB. 

With such a weight, the following stresses result: 

6.75 x tana — 5.48 tons. 

(AB)" = — (de) n — — 5.48 tons. 

(DE)" — — (ef)" = — 10.97 “ 

(EF) " = - (f g y = - 16.45 “ 

(EG) " = - (gJi)" = - 21.94 “ 

(GH)" = — (bh)" = — 27.42 “ 

(BH)" — — (27.42 + 6.75 tan ( 3 ) = — 30.37 tons. 

From these results are to be selected the greatest chord 
stresses. 

For examples: 

Resultant (GH) = — (35.99 + 75.00) = — 110.99 tons. 

“ {FG) = - (21.94 + 47 - 44 ) = ~ 69.38 « 

In short, precisely as the operation has been done before. 

The resultant web stresses caused by this negative reaction, 
are: 

(Ad) = (3 + 6.75) x sec a = + 12.58 tons. 

(dD) — — (3 + 2.73 + 6.75) = — 12.48 tons. 

These are the “ singular ” stresses already mentioned. 

Collecting and arranging the results, the following resultant 
stresses are obtained : 


(dE) 

= + 

23.19 

tons. 




(Ad) 

= + 

12.58 

< < 

(Ad) = 

- 53-78 

tons. 

(dD) 

= + 

24.50 

u 

(dD) = 

— 12.48 

it 

m 

= + 

28.17 

u 

(eE) = 

~ 24.57 

tt 

W) 

= + 

49.19 

u 

i/E) = 

— 40.86 

u 

m 

= 4. 

75-19 

«< 

igG) = 

— 61.02 

n 

(GA) 

= + 

105.50 

u 

(AH) = 

- 84.51 

u 

(Hb) 

= + 

117.64 

u 

(bB) = 

- 217-33 

tt 


(AD) 

= — 

7.92 

tons ; 4- 33.87 tons. 



(DE) 

= - 22.13 

“ ; + 33-87 “ 



200 


SWING BRIDGES. 


( EF ) = — 42.61 tons; + 35.25 tons. 


(FG) =- 69.38 
(GH) = — 110.99 
{BH) = - 153.82 
ide) — + 7.92 

(ef) =4- 22.13 
(fg) = + 42 . 6 i 
(gh) = +. 69.38 
(Jib) — 4- no.99 


“ ; + 2.76 “ 

<< 

<< 

“ ; — 45.62 tons. 
“ ; — 45.62 “ 

“ ; - 35-25 “ 

“ ; — 2.76 “ 


The same stresses exist, of course, for corresponding mem¬ 
bers in the arm BC. 

It is thus seen that the portions dh, d'Ji, AG, and CG', of 
the chords must be counterbraced. Ad, Cd', dD, and d'D', 
only, of the web members need the same treatment. 

It may happen that, with a moving panel load at D, the 
reaction at A will be negative, in the search for main web 
stresses. In such a case the method of operation is simply 
an extension of that used in the example. If the numerical 
value of this negative reaction is equal to, or less than, a 
moving panel load (which may rest at A), a weight equal to 
this reaction is to be taken as hung from A, and the panel 
load (moving) at D is to be taken as hung from that point, 
while the arm AB is to be considered as an overhanging one. 
If the reaction, however, is greater than a moving panel load, 
then two such loads are to be taken as hanging from A and 
D with the overhanging condition of the arm. 

A whole panel moving load is taken at A for prudential 
reasons. If the load were of uniform density, then a half 
panel moving load would be taken at A. 

Negative reactions by the formula for any number of mov¬ 
ing panel loads near the end are to be treated in exactly the 
same way; for it is to be remembered that negative reactions 
in an actual truss, in this case, cannot exist. 

It may be urged that the case of partial continuity, taken 
in the preceding Article, should be treated according to the 
principles developed in this, by taking the middle span equal 
to zero. 



ENDS SIMPLY RESTING ON SUPPORTS. 


201 


Making such an assumption, however, would be a depart¬ 
ure from the real state of the truss. The safe way would be 
to determine the greatest stresses by both methods, and select 
the greatest of the two sets of results. 

Differences would be found only in the upper chord ten¬ 
sion, lower chord compression, and main web stresses. 

In the present case, if there are two or more systems of 
triangulation, each is to be treated precisely as the example 
has been. 

. This case really includes that of a centre-bearing turn-table 
with two points of support at the centre, as shown in the 


7i 7i' 



figure. HH' is free to “ rock ” on the central point B, and as 
the motion is always very small, BH (horizontal distance) is 
essentially equal to BH' (also horizontal) at all times. From 
this it results that the reaction at H will always be equal to 
that at H', consequently the diagonals Hh' and H'h must be 
introduced. 

Now as HH' is really a part of the truss, attached to and 
moving with it, the whole bridge, AC , is simply a continuous 
truss of two spans supported on the fixed point B. All the 
conclusions and formulae, therefore, of this Article, apply to 
it directly. R 2 will be the reaction at B , and M will be the 
moment over the same point. 

According to the principles established, Hit will receive its 
greatest stress when AB, only, carries moving load. Since 
the pressure on H is always equal to that on H' also to a 
half of the reaction at B, there results: 

(Hh') = X ^ hHh ; 

% 

in which R\ is the reaction at B due to the moving load on 
















202 


SWING BRIDGES . 


AB only, considered as a simple truss. The greatest stress 
in H'h is, of course, equal to ( H/i ). 

The greatest stress in Hh (equal to (H'h)) is found, as 
before, by putting the moving load on BC and AG. 

No locomotive excess has been taken, but precisely the 
same conditions of loading hold whether such excess is taken 
or not. 

It will only be necessary to remember that the locomotive 
may be at either end of the train, and that the greatest results 
arising from the two positions are to be selected. 


CHAPTER V. 


SWING BRIDGES. ENDS LATCHED TO SUPPORTS. 


Art. 39.—General Considerations. 


It has already been stated that the object of fitting the 
ends of a swing bridge with a latching apparatus is to enable 
those ends to resist a negative reaction, or in other words, to 
prevent their rising from the points of support. All “ham¬ 
mering” of the ends will thus be prevented. 

It has further been shown in the preceding Chapter that if 
there are ahuays two points of support at the center, for each 
system of triangulation, the ends will never tend to rise. It 
was also observed in the preceding Article that with a pivot, 
or centre-bearing turn-table, the bridge always presents the 
case of continuity with two spans only, whatever may be the 
number of apparent points of support at the centre. 

In this chapter, then, it will only be necessary to consider 
the one case of continuity with a single point of support 
between the extremities of the bridge. 

Art. 40.—Ends Latched Down—One Point of Support Between Extremi¬ 


ties of Bridge—Example. 


The general formulae required in this case are Eqs. (i), (2), 
(3), and (4), of Article 38, and they are here reproduced. 



j hp{P - £)z + 2 P(i 2 - P)z 




(2). 



204 


SWING BRIDGES. 


R 2 = I | 2 Pg + 2 Pz -2M\ . (3). 

^? 3 = I | ZP(f-z) + M} . (4). 


These involve the condition 4 = l 2 = /, which will appear in 
the example. 

If this condition does not exist in any case, the formulae to 
be used are Eqs. (io), (n), (12), and (13) of Article 35, but 
they are to be used in precisely the same manner as will be 
Eqs. (1), (2), (3), and (4). 

This case was essentially treated in the preceding Article, 
insomuch that with ordinary moving loads precisely the same 
conditions of loading, for the greatest stresses, are required in 
the two cases. The results themselves, however, will be dif¬ 
ferent for the upper chord compression, lower chord tension, 
and counter stresses. 

It will probably be as expeditious and labor saving, never¬ 
theless to find the reactions and chord stresses due to each 
moving panel load, and then combine the results thus found 
with those due to the fixed load alone, in the usual manner. 
Such is the method to be used in the example. 

Since Eq. (i) shows that M is always negative, Eq. (2) 

1 

shows that with a given value of 2 P (/ — z), R 1 will have its 
greatest positive value when no moving load is upon the span 
4. The expression for the shear in any counter: 

s = R^ — n'w — n (W + W'); 

(in which ri is the number of moving loads between R 1 and 
the counter, and n the number of fixed loads similarly lo¬ 
cated), will have its greatest value for n' — o. Hence, for the 
greatest stress in any inclined counter, the moving load must ex¬ 
tend from the centre to the foot of the counter in question. This 
is precisely the condition used previously. 

As usual, the stress in the vertical which cuts the upper 




ENDS LATCHED DOWN. 


205 

extremity of the counter must be found, for it may be the 
greatest in that member. 

Precisely the same reasoning used previously shows that 
the greatest stress in any main web member ( inclined ) exists 
when the moving load covers the whole of one span , and that 
portio?i of the other included between the free end and the foot of 
the member considered. 

The stress in the vertical which cuts the upper extremity 
of the inclined web member is to be found with the same 
condition of loading ; it will usually be the greatest possible. 

The truss to be taken for an example, and all the data, are 
exactly the same as those used in Article 38. The figure and 
the data are reproduced below: 


d e f a h b H a' &• d’ 



Total length = AC = 2/ = 2 AB = 2BC = 144 feet. 

Uniform depth = dD — bB = 16 feet. 

Panel length = AD — DE = etc. = 13 feet. 

BH=BH'= 7 feet. 

Total fixed weight per foot = 1200 pounds (nearly). 

Upper chord panel fixed weight = W = 2.73 tons. 

Lower chord panel fixed weight = W' = 5.00 tons. 

Uniform panel moving load — w— 19.50 tons. 

The inclined web members, except the end posts, are for 
tension, and the verticals for compression. 

The moving load traverses the lower chord. The weight 
of the floor system is taken at about 350 pounds per foot of 
track. The fixed weight at A will be taken at three (3) tons, 
and that at b at 1.47 tons. Full panel loads of both kinds 
will be taken at H and H'. 

For a single moving panel load on the arm AB: 


M=- J (/*-/)* 


(5). 


















20 6 


SWING BRIDGES. 


R x = i | w (/— z) + Mj . . . . (6). 




The distance z is to be measured from A or C according as 
the arm AB or CB is considered. 

The trigonometrical quantities used in this example are 
the same as those employed in the preceding Article. They 
are the following: 

tan AdD — tan a — 0.8125 
sec “ —sec a — 1.29 
tan HbB — tan (5 — 0.4375 
sec “ —sec fi = 1.09 


The following quantities are also taken from the example 
in the preceding article: 


z 

= 

. . 1 — z 

= 59 

• ( /2 

-«*)« = 

65195.OO 

z 

= 26 

. . 1 — z 

= 46 . 

• (C 

— z 2 )z = 

I 17208.00 

z 

= 39 

. . 1 — z 

= 33 • 

■ [C 

- z t )s = 

142857.00 

<** 

/J 

= 52 

. . 1 — z 

= 20 . 

■ (C 

_ r £\ r , _ 

128960.00 

Z 

=65 

. . 1 — Z - 

= 7 • 

■ {C 

_ r 2 \„ _ 

/O - 

62335.OO 



I 

Al ' 1 

1 

20736 

— 0.00004823. 




0.00004823 X 

19.5 = 0. 

00094 . 




w 

19.5 

= O.271 





7 

72 

• 


By 

using 

these quantities in 

Eqs. (5) 

1, (6), and (7): 

w at 

D . 

. Bi — + 

15.14 tons . 

• — 

— 0.851 tons. 

w “ 

E . 

. — T 

1094 

(( 

• — 

- i -53 

w “ 

F . 

. 7 = + 

7.07 

u 

• = 

- 1.87 

w “ 

G . 

. 7 = + 

374 

u 

• -^3 = 

— 1.68 “ 

w “ 

H . 

. -7 — + 

1.09 

<< 

. — 

— 0.81 “ 



= + 

37-98 


2^3 = 

— 6.741 “ 









ENDS LATCHED DOWN. 


20 / 


The greatest negative reaction at the extremity of one arm 
will exist when the whole of the other is covered by the mov¬ 
ing load, and its value is seen to be — 6.741 tons. The resist¬ 
ance of the latching apparatus must be sufficient to oppose 
this with a proper safety factor. 

Under the same circumstances, with ends not latched down, 
it was found that the reaction 2XA was 44.69 tons (see Article 
38); but 37.98 + 67.41 = 44.721 tons, which is essentially 
equal to 44.69 tons, as it should be. 

Counter Stresses. 

dE is the only counter needed, since with moving loads at 
E, EG, H , the reaction at A is: 

R x — 10.94 4 - 7.07 + 3.74 +1.09 = + 22.84 tons ; 
consequently 2 P= o at E. 

The shear, or vertical component of the stress, in dE is: 
s= 22.84 — (3 + 2.73 + 5.00) = 12.11 tons. 

Hence, (dE ) = s x sec a = + 15.62 tons. 

With the moving load covering DH, dD acting as a counter 
will sustain a tensile stress equal to 

(dD) — + (19.5 + 5.00) = 4- 24.5 tons. 

With the same condition of loading, Ad receives its great¬ 
est compressive stress : 

(Ad) — —R\ x sec a — — 37.98 sec a — — 48.99 tons. 

Main Web Stresses. 

The main web stresses are found precisely as in Article 38, 
and there is no need of repeating the operation here. 

The values of the stresses will be reproduced in the proper 
place. 

Chord Stresses. 

The chord stresses due to the fixed load alone are the same 
as those determined on page 195 of Article 38. They will 


208 


SWING BRIDGES. 


not be reproduced, but references will be made to them as 
they are. 

Those caused by the moving load alone will be determined 
by placing a panel load at each panel point successively, 
and finding all the chord stress in both arms due to it, then 
tabulating the results, and, finally, combining them in the 
manner already shown in several instances. 

The counter dE will be supposed to come into action for 
the weights E, E, G, and H. 

The panel loads at F, G, and H will cause apparent com¬ 
pression in some, or all, of the inclined members Ef, Eg , and 
Gh. The resultant action of fixed and moving loads in those 
members, however, will in all cases be tension. 

The detailed expressions for the chord stresses due to one 
moving panel load only will be given, as all the others are 
like it. For this purpose take w at F. 

R\ — + 7.07 tons ; R 3 = — 1.87 tons. 

w — R x — 19.5 — 7.07 = 12.43 tons. 

{AD) = {DE) = + R 1 tan a — + 5.74 tons. 

{de) — (ef) = — 2 x R 1 x tana-==. — 11.49 tons. 

(EF) = - (fg) = + n.49 + 5-74 = + 17.23 tons. 

(FG) ~ — (gh) = 17.23 — 12.43 x tana = + 7.13 tons. 

(GH)= — (/iff) = 7.13 — 12.43 x tan a = — 2.97 tons. 

(■ HB) = — 2.97 — 12.43 x tan ft = — 8.41 tons. 


(CD) = 
(DE') = 
(E'F) = 
lF'G') = 

(1G'H') = 
(HD) = 


( d ' e ) 

Vf) 

t fg ') 

(£'*’) 
(h'b) -- 

7.60 — 


= i?, x tan a 


— 2 X 


u 


u 


= 3 X 

— 4 x “ 
R 3 x tan 


u 

u 

<< 

u 


— 1.52 tons. 

— 3.04 “ 

— 4.56 

— 6.08 

— 7.60 

— 8.42 


u 


u 


u 


u 


The following checks by moments should be observed. 
Moments about b give : 


(HB) 


. R \ x 72 - 19-5 x 33 

l6 


— 8.40 tons ; 



^3 X 72 

16 


= — 8.415 tons. 




ENDS LATCHED DOWN. 


209 


Moments about /give: 

, ^1 X 39 

(• EF) = l6 -- * = + I7-23 tons. 

The following tables are found by following the same 
operation for all the weights. 



( de ) 

(*/) 

( fg ) 

( gt ) 

w 

w at D 

— 12.3 

— 8.76 

— 5-22 

- 1.68 

+ i. 86 

“ “ E 

- I 7-73 

- I 7-78 

— 10.82 

- 3-87 

+ 3-09 

“ “ j ? 

- 11.49 

- n -49 

- 17-23 

- 7-13 

+ 2.97 

“ “ G 

— 6.08 

- 6.08 

— 9.I2 

— 12.16 

+ 0.65 

“ “ H 

- i -77 

- i -77 

- 2.66 

- 3-55 

- 4-42 

/ 



( AD ) 

( DE ) 

( EE ) 

( FG ) 

( GH ) 

( HE ) 

w at D 

+ 12.3 

+ 8.76 

+ 5-22 

+ 1.68 

— 1.86 

- 3-77 

“ “ E 

+ 8.89 

+ 8.89 

+ 10.82 

+ 3-87 

- 3-09 

- 6.94 

“ “ F 

+ 5-74 

+ 5-74 

+ 17-23 

+ 7-13 

- 2.97 

— 8.41 

“ “ G 

+ 3-04 

+ 3-04 

+ 9.12 

+ 12.16 

— 0.65 

- 7-54 

“ “ H 

4- 0.89 

+ 0.89 

+ 2.66 

+ 3-55 

+ 4-42 

- 3-63 



( d ' e ) 

«/') 

(//) 

( g ' V ) 

( h ' b ) 

w at D 

+ 0.69 

+ 1.38 

+ 2.07 

+ 2.76 

+ 3-45 

“ “ E 

+ 1.24 

+ 2.49 

+ 3-73 

+ 4-97 

+ 6.22 

“ “ F 

+ 1.52 

+ 3-04 

+ 4.56 

+ 6.08 

+ 7.60 

“ “ G 

+ 1.37 

+ 2.70 

+ 4.10 

+ 5.46 

+ 6.83 

“ “ H 

+ 0.66 

+ 1.32 

+ i -97 

+ 2.63 

+ 3-29 


14 










































210 


SWING BRIDGES. 



( CD ') 

( DE ) 

( E ' F r ) 

( EG ) 

( G ' H ) 

(HB) 

tv at D 

- O.69 

- 1.38 

— 2.07 

— 2.76 

- 3-45 

— 3.82 

“ “ E 

— 1.24 

- 2 49 

- 3-73 

- 4-97 

— 6.22 

- 6.88 

“ “ p 

- 1-52 

- 3-04 

- 4-56 

- 6.08 

— 7.60 

— 8.42 

“ “ G 

- 1-37 

- 2.73 

— 4.10 

- 5-46 

- 6.83 

- 746 

“ “ H 

— 0.66 

— 1.32 

- 1-97 

— 2.63 

- 3.29 

- 3 64 


Using the main web stresses and the fixed weight chord 
stresses found in Article 38, the following greatest stresses at 
once result: 


(dE) = + 
(Ad) = + 
{dD) = + 
(De) = + 
(Ef) = + 
(Fg) = + 


15.62 tons. 
12.58 “ 
24.50 “ 
28.17 “ 

49.19 “ 

75.19 “ 


(Ad) = - 
(dD) = - 
{eE) = - 

(fF) = - 
(gG) = - 
(hH) = - 

m = - 


48.99 tons. 
12.48 “ 

24.57 “ 

40.86 “ 

61.02 4< 

84.51 u 

51 7*33 “ 


(GJi) = + 105.50 
{Hb)~ + 117.64 


{AD) = — 7.92 tons ; 

{DE) — — 22.12 “ ; 
{EE) = - 42.59 « ; 

(EG) = - 69.34 “ 

{GH) — — 110.96 “ 

{ BH )= - 153.83 “ 
(de) = + 7.92 “ ; 

(ef) = + 22.12 u ; 

(fg) = + 42.59 “ ; 

(^) = + 69.34 “ 

{kb) — + 110.96 “ 


+ 28.42 tons. 
+ 16.16 “ 

4- 18.89 a 


- 46.98 tons. 

- 3472 “ 

- 18.89 “ 


The web members dD, d'D', Ad, Cd', and the portions AF, 
CF'j dg, d'g' of the chords need counterbracing. 

















ENDS LATCHED DOWN. 


211 


The chord stress (GH) = — 110.96 tons requires the mov¬ 
ing load to cover BC and AG. 

With the moving load on AB only, there is some ambi¬ 
guity in the stresses (de), compression , and ( DE ) tension. 

In such a case the reaction R x is 37.98 tons, and the web 
member De may be neglected. Under such an assumption, 
by taking moments about d and E successively, there will 
result: 

(AD) = (DE) = g X 13 = + 28.42 tons; 


(de) = (ef) = - W ~ 3) x 26 - 27.23 x 13 


— 34.72 tons. 


This ambiguity cannot be avoided if both the web mem¬ 
bers dE and De exist. It might also have been noticed in 
the case last treated. 

It has already been noticed that the downward reaction of 
6.74 tons must be resisted by the latching apparatus. 

If there are two or more systems of triangulation, the pre¬ 
ceding principles hold true for each. Also, if there is loco¬ 
motive excess, precisely the same methods are to be em¬ 
ployed. 

The observations which were made at the end of Article 38 
on a pivot or centre-bearing turn-table, over which there are 
two points of support for the truss, apply, exactly as they 
stand, to this case. The value of R' 2 must, however, be found 
by Eq. (3) of this Article. 





CHAPTER VI. 

SWING BRIDGES ENDS LIFTED. 

Art. 41.—General Considerations. 

In the preceding chapter there was noticed, in detail, the 
method of prevention of “ hammering,” by latching down the 
ends of a swing bridge of two spans. It was also there 
noticed that the necessity of such an arrangement could only 
exist in the case of continuity with two spans. For precisely 
the same reasons given in connection with that case, the neces¬ 
sity of lifted ends can exist in the event of continuity with two 
spans only. 

It is plain that if the ends of a swing bridge are pressed 
upward by forces exceeding the greatest negative reactions 
determined for latched ends by the formulae of the last chap¬ 
ter, there can be no hammering, for the ends can never leave 
their seats or supports. 

By a proper device, then, the ends should be pressed up¬ 
ward by forces at least equal to the negative reactions deter¬ 
mined for latched ends. 

In order to provide for any contingency, however, which 
may arise, the upward force should somewhat exceed such a 
value. 

Art. 42.—Ends Lifted—One Point of Support Between Extremities— 

Example. 

The figure represents the truss to be taken as an example. 
The span, depth of truss, and panel lengths, excepting hh\ 





212 







ENDS LIFTED. 


213 


are the same as those taken in the two preceding cases; the 
loading is also the same. 

The following are the data to be used : 

AC =2^Ah+^\= 144 feet. 

Uniform depth of truss = 16 “ 

Panel length =13 “ 

Ah’ =14 “ 


Uniform fixed upper chord panel load — W — 5.00 tons. 

“ “ lower “ “ “ = W'= 2.73 “ 

“ moving “ “ = w = 19.50 “ 

Moving load for unit of length = 1.50 “ 


The truss is a deck one, as the moving load passes along 
AC; and as the figure shows, there are two systems of tri¬ 
angulation. It will be assumed, though not strictly true, that 
the same panel loads are found at h and h! as at the other 
panel points. 

Let the inclination of Gh to a vertical line be denoted by a. 
“ “ BA “ “ “ 0 . 

“ “ Be; “ “ 11 6. 


Then tan a — 0.8125 ; 
“ “ = 0.4375 ; 

“ “ 6 = 1.25 ; 


sec oc = 1.29; 
“ 0 = 1.09; 
“ d = 1.6. 


Each system of triangulation is to be treated as an inde¬ 
pendent truss. The fixed weights at D and A will be taken 
as belonging to the system ADeF\ etc., while that atV will be 
assumed to belong to the other system. Similar observations 
apply to the other arm. As in the preceding case, a fixed 
load of three (3) tons will be taken at A or C. 

The stresses in a swing bridge with ends lifted may be 
considered as composed of the stresses in two other trusses, 



214 


SWING BRIDGES . 


one with ends latched down and subjected to the same loads 
(both fixed and moving), and the other subjected to the 
action of the upward pressures only, at the ends; the dif¬ 
ferent trusses being supposed of the same form and dimen¬ 
sions in all their parts. 

From this, it at once follows that the positions of the mov¬ 
ing load for the greatest stresses (when the ends are lifted) 
are exactly the same as those determined in the preceding 
chapter. 

For the stresses in the counters, then, or for those in the mem¬ 
bers which slope downward from the upper chord arid toward 
the ends, the moving load must extend from the centre to the 
upper extremities of such members. 

These stresses will be compressive, and the member in 
which such stress is first found is to be determined in the 
manner already shown. 

In order to find the greatest tompressive stress in any web 
member, in one arm, sloping downward from the upper chord, 
and toward the centre, the moving load must extend from the 
end of that arm to its upper extreniity, and at the same time 
cover the whole of the other arm. 

These conditions of loading are to be taken while the ends 
are lifted, but it will also be necessary to find the web stresses 
for the open draw in the vicinity of the end, as some of these 
will be the greatest stresses in the web members there located. 

It is to be borne in mind that any two web members which 
intersect in that chord which does not carry the moving load, 
take their greatest stresses together. 

Although positions of moving load for the greatest chord 
stresses may be assigned, it will probably be the shortest and 
most labor-saving method to find the chord stresses due to 
the fixed load and upward pressure together, then find those 
due to each moving panel load alone, and combine the results. 
This method will be used. 

The example will now be treated. 

The following quantities are determined on the supposition 
that the ends are latched down, by Eqs. (i), (2), and (4) of 
Article 40. 


ENDS LIFTED. 


215 


2=13 feet 

2 = 39 “ 

2 = 65 “ 


System ADdEf\ etc. 

M— — 61.28 . . R x — + 15.14 . . E.1 = —0.851 tons. 

M — — 134.29 . . R t = + 7.07 . . A*., = — 1.87 “ 

M — — 58.59 . . = 4 1.09 . . R 3 — — 0.81 “ 


- 3.531 


i i 


z — 26 feet 


2 = 52 


i 4 


System ADeFg , etc. 

M = — 110.18 . . R x = + 10.94 . . R a = — 1.53 tons. 
M= — 121.22 . . R l = 4 3.74 . . R 3 = — 1.68 “ 


Each of these results, it is to be observed, is for a single 
panel moving load placed at the panel-point denoted by the 
value of z. They have been used in the two preceding cases. 

The chord stresses due to each panel moving load alone 
will first be found. As these are all found by exactly the 
same method, the detailed expressions for two only (one in 
each system) will be given. 


System ADdEf, etc. 

Panel moving load at f : 

z — 39 ^et; R t = + 7.07 tons ; E 3 = — 
(Ad) 

m 

(. fk) 

(hti) 

(DE) 

(EG) 

(GB) 

(Cd') 

( d'f) 

(f’/i) = - 4 

(h/i) = 6.08 — R 3 ( tan a + tan ( 3 ) 

(. DE') = - (Cd') 

(E'G') = + 3 R$tan a 
(G'B) = + 5 “ “ 


= — R x tan a 
= — 2 Ri tan a 

— — 11.49 + (12.43 — 7.07) tan a 
= — 7.20 + 12.43 (tcm a + tan ( 3 ) 
= - ( Ad ) 

= 5.74 4- 2 R x tan a 

— 17.23 — 2 x 12.43 x ian a 

— — R 3 tan a 




u 


u 


u 


u 


— — 

1.87 

tons. 

= — 

574 

u 

= — 

11.49 

<« 

= — 

7.20 

u 

1= + 

8.30 

u 

= 4 

574 

a 

= + 

17-23 

u 

= — 

2.96 

u 

= + 

1.52 

u 

= 4- 

3-04 

u 

= + 

6.08 

u 

= 4" 

8.42 

u 

— — 

1.52 

u 

— — 

4.56 

it 

— — 

7.60 

{( 




216 


SWING BRIDGES. 


As numerical checks, the moment method gives the fol 
lowing results: 


(G£) = 


R\ x 65 — 19.5’ x 26 
16 


— 2.96 tons. 


(hh) = - R * * 72 = + 8.41 tons. 


(G'i?) = R * ^ ^ = — 7.60 tons. 


System ADeFg, etc. 

Panel moving load at e : 

£=26 feet; ^ = + 10.94 tons ; R 3 
(Ae) = — R t tan a 
(eg) = (Ae) — 2.38 x to a 
(gg') — (eg) + (19.5 — Ri) (tan a + tan d) 

(DF) =2 R t tan a 
(FB) = ( DF) — 2 (19.5 — R\) tan a 
( Ce ') =— R 3 tan a 

(e'g') = ~ 3 “ “ 

(g'g) = (e'g') — R z (tan a + tan 6) 

(DF') — 2 R 3 tan a 
(F'B) = 4 


— 1.53 tons. 

— 8.89 “ 

— 10.82 
-+- 6.84 
+ 1778 


+ 

+ 

+ 

+ 


u u 


3- 87 

1.24 

373 

6.88 

2.49 

4 - 97 


ft 


it 


(( 


u 


a 


u 


u 


It 


u 


Moments give: 


(FB) 


R 1 x 52 — 19.5 x 26 _ 
16 — 


-f 3.86 tons. 


t~)\ x ^2 

(F B) = —- 3 = — 4.97 tons. 







ENDS LIFTED. 


217 

All the results for the two systems give the four tables 
below: 



(Ad) 

(df) 

(A) 

(hh) 

(■ de ) 

(EG) 

(GB) 

w at d 

— 12.30 

~ 8.76 

- 1.68 

+ 3-77 

+ 12.30 

+ 5+2 

- t.86 

w at f 

- 5-74 

- 11.49 

— 7.20 

+ 8.30 

+ 5-74 

+ 17.23 

— 2.96 

w at h 

— O.89 

- 1.77 

- 3-54 

+ 3-63 

+ 0.89 

+ 2.66 

+ 4.43 



( Cd ) 

{ df ) 

(/*) 

(hh') 

( DE ) 

( E ’ G ) 

( GB ) 

w at d 

+ O.69 

+ 1.38 

+ 2.77 

+ 3.83 

— 0.69 

— 2.07 

~ 3.46 

w at f 

+ 1.52 

+ 3.04 

+ 6.08 

+ 8.42 

— 1.52 

- 4.56 

— 7.60 

w at h 

+ 0.66 

+ 1.32 

+ 2.63 

+ 3-64 

— 0.66 

+ 1-97 

~ 3.29 



(Ae) 

• k?) 

{gg) 

(DF) 

(FB) 

w at e 

— 8.89 

— 10.82 

+ 6.84 

+ I 7-78 

+ 3-87 

“ “ E 

- 3-04 

— 9.12 

+ 7-54 

+ 6.08 

+ 12.16 



(Ce) 

kg) 

{gg) 

(DF') 

(F'B) 

w at e 

+ 1.24 

+ 3-73 

+ 6.88 

~ 2-49 

- 4-97 

“ “g 

+ i -37 

+ 4.10 

+ 7-56 

“ 2.73 

- 5-46 


The open draw stresses due to the fixed weight alone are 
the following: 

3 x tan a = 2.44 tons. 3 x tan d = 3.75 tons. 

W x “ =4.06 “ W x “ —6 .25 “ 

W' x “ = 2.22 “ W' x “ = 3.41 “ 


























































218 


SWING BRIDGES . 


IF x tan (5 — 2 . 19 tons. 
IF' x “ = 1.19 “ 


(Ad) 

= 4- 3 x tan a 

— 

+ 2.44 

tons. 

(de) 

— (Ad) 4- W tan a 

— 

+ 6.50 

a 

(ef) 

— (de) 4- \2 (3 4- IV') 4- W\ tan a 

— 

4- 19.88 

u 

(fg) 

— ( e f) + \2 (W + W') + W\ tan a 

— 

+ 36-50 

u 

(gt) 

— (fg) 4- (3 4- W 4 2 W') (tan a + tan d) 

+ 

W tan d 




— 

+ 70.51 

tons. 

(hh) 

— (gti) + 2 (JV + W') (tan a 4- tan ( 3 ) 

4- 

W ta?i (5 




— 

+ 92.02 

tons. 

(DE) 

= — (2x34- W ) tan a 

— 

— 7.10 

u 

(EF) 

= (DE) — 2 W tan a — W tan a 

— 

- 17-44 

it 

(FG) 

= (EF) — 2(3 4- W+ W) tan a— W' tan a 

— 

-37-io 

it 

(GB) 

= (EG) — (4 W 4- 3 W') tan a 

— 

— 60.00 

it 

As 

a numerical check: 




(GB) 

— (3 W + 2 W') tan (5 — (3 + 2 W + 2 W' 

) tan d = 



— 92.02 tons = — (hh). 

Again, by moments: 


//./^ _ 5 x 7-73 X 33 + 3X 72 -7X 2.73 
(M) - ^ 


4- 92.02 tons. 


The chord stresses resulting from the upward pressure 
alone still remain to be found. 

The total negative reaction, supposing the ends to be 
latched down, has been shown to be — (3.53 4- 3.21) = — 6.74 
tons. A margin of safety, however, of two tons will be taken ; 
i. e ., it will be assumed that the total upward pressure at each 
end of the bridge has a value of 8.74 tons. 

In the example, and in all cases where two or more systems 
of triangulation have a common point of support at the ends, 
some ambiguity necessarily arises in regard to the upward 
pressure. The proportion of the excess carried by either sys¬ 
tem is indeterminate; and if there is no excess, the propor¬ 
tion of the upward pressure carried by either system, during 
partial loading of one or both arms, is also indeterminate. 



ENDS LIFTED. 


219 


In the absence of anything better, it will be assumed that 
the excess, in the example, of two tons is equally divided be¬ 
tween the two systems. It will farther be assumed that the 
upward pressure, under all circumstances of loading, is 4.53 
tons for the system ADdEf, etc., and 4.21 tons for the system 
ADeFg, etc. The chord stresses due to the upward pressures 
will then be the following: 

(Ad) = — 8.74 X tana 
(de) — (Ad) — 4.53 x tan a 
(ef) — (de) — 2 x 4.21 x tana 
(. fg ) '=(</) - 2 X 4.53 
(g/i) = (fg) — 4.21 (tana + tan 3 ) 

(hh f ) (g/i) — 4.53 (tan a 4- tan fi) 

(DE) — 2 x 4.21 x tan a 4 - 4.53 tan a 
(EF) = (DE) + 2 x 4.53 x tan a 
(FG) = (EF) 4 - 2 x 4.21 
(GB) = (FG) 4 - 2 x 4-53 

As numerical checks: 

/7 8.74 x 72 

(hh ) =- {$ ■ '- = — 39-33 tons. 

(GB) -f- 4.53 x tan + 4.21 x tan 3 — 39.32 tons = — (hh). 

The stresses in the web members, existing with a passing 
load, will next be found, and those which may be termed 
counter stresses will first receive attention. 

System ADdE, etc. 

Moving loads at f and h\ 

R x — 7.07 + 1.09 + 4.53 = + 12.69 tons. .*. 2P = o at f 
and (f£) will be the first counter stress. 

The shear, in fE, is: 

j = 12.69 — 7.73 = 4.96 tons. 

(fE) — — s x see a = — 6.40 tons. 

.-. (dE) = (s + 2.73) sec a — + 9.92 tons. 


= — 7.10 tons. 
= — 10.78 “ 

= — 17.62 “ 

= — 24.98 “ 

= - 33-66 “ 

= - 39-32 “ 

= -f IO.52 “ 

= 4 - 17.88 “ 

= 4- 24.72 “ 

= 4 - 32.08 “ 



220 


SWING BRIDGES. 


Moving loads at d, f and h : 

R 1 — 12.69 + 15.14= 4 - 27.83 tons. 
s (for AD) = 27.83 tons. 

.*. (AD) — y s x sec a — y 35.9 tons. 
(dD) = — s — — 27.83 tons. 


System ADeE, etc. 

Moving loads at e and g: 

R t = 10.94 + 3-74 + 4.21 = 18.89 tons. .*. 2 P = o at e, and 
(De) is the first counter stress. 

^ (for De) = 18.89 ~ (3 + 2.73) = 13.16 tons. 

.\ (De) = — s x sec oc = — 16.98 tons. 

.-. (AD) = (R x — 3 ) sec a = + 20.86 tons. 

The main web stresses existing with the moving load are 
found as follows: 

System ADdE, etc. 

Moving load on Ch' and at d : 

R\ = 1 5-!4 + 4-53 - 3-53 = 16.14 tons. 

(dE) = — (19.5 + 5 — 16.14 )seca= — 10.78 “ 

(Ef) — — (dE) y W' sec a = + 14.30 “ 

Moving load on Ch and Af: 

R t = 16.14 + 7.07 = 23.21 tons. 

(fG) — — (2 W + 2 w + W' — R x ) sec a — — 36.79 “ 

(G/i) — — (fG) + W' sec a = 4- 40.31 “ 

Moving load on Ch and Ah. 

R x — 23.21 4 1.09 = 24.30 tons. 

(hB) — — (3 W y 3 w + 2 W' — RJ sec ft = — 59.58 tons. 


ENDS LIFTED . 


221 


System ADeF, etc. 

Moving load on Ch' and at e: 

R\= 10.94 + 4.21 — 3.21 — 11.94 tons. 


(eF) = — ( W 4 - w 4 - W' 4- 3 — R\) sec a = — 23.59 tons. 
(Fg) — — (eF) 4 W' sec a =4 27.11 “ 

Moving load on Ch' and eg: 

R 1 — 11.94 4 3.74 = 15.68 tons. 

(gB ) = — {2 ( W 4 W 4 - w) 4 3 — R x \ sec d = — 66.85 tons. 


A few of the open draw web stresses are the following 


( AD ) = — 3 sec a — — 3.87 tons. 

(dD) =0 — 0.00 

(dE) — — Wsec a = — 6.45 

(De) = 4 (3 4 W) sec a =4 7.39 

(eF) = — (34 W+ IV) sec a — — 13.84 

(Ef) — + (W + W') sec a = 4 9.97 


u 


u 


a 


u 


u 


It is unnecessary to give others, as they are not needed; 
only two of these, it will be seen, are used. 

All of the greatest stresses in the truss may now be written 
by the usual method of combining the results for the different 
cases of loading. 

They are the following: 


(AD) = 

- 3-87 

tons; 

+ 

56.75 

tons. 

{dD) = 

- 27.83 

a 




(dE) = 

— IO.78 

u 

+ 

9.92 

u 

{De) = 

— I6.98 

u 

4 - 

7-39 

u 

{eF) = 

“ 23.59 

u 




(Ef) = 

— 6.4O 

u 

+ 

14.30 

a 

(fG) = 

- 36.79 

(( 




(Fg) = 



+ 

27.1 I 


(gB) = 

— 66.85 

(( 




(Gh) = 



+ 

40.31 

a 

{hB) = 

- 59-58 

(c 





222 


SWING BRIDGES. 


(Ad) = + 

2.44 tons; 

- 35.52 tons, 

(de) = + 

6.50 

« 

- 38-23 

u 

(ef) = + 

19.88 

u 

- 3970 

a 

(.fg) = + 

36.50 

(i 

— 20.84 

u 

(gh) = + 

77-15 

a 



(Mi) = + 

112.51 

u 



(DE) = - 

7.10 

<< 

+ 46.21 

a 

(EF) = - 

17.44 

<< 

+ 4941 

u 

(FG) = - 

37.10 

<< 

+ 38.76 

a 

(GB) = - 

57*52 

u 




The web members AD, dE, De, Ef, and the portions Ag 
and DG of the chords must be counterbraced. The same 
treatment must of course be given to corresponding members 
and portions in the other arm. 

The particular form of truss in the figure has been so chosen 
as to illustrate faults of designs, in general, in consequence of 
possible ambiguity in the stresses. 

If possible, ambiguity should always be avoided. In the 
present case it would have been far better to have had one 
system of triangulation, and supported the chords by light 
verticals, designed to resist compression, extending from the 
apices. 

Precisely the same methods of loading and treatment would 
be used if there were two apparent points of support above B, 
that point still existing as the real point of support of the 
truss. In fact, the same general observations as those which 
were made in the last portions of Articles 38 and 40 apply in 
this case also. 

The same methods of loading and treatment would also be 
used if there were locomotive excess, or if there were one or 
more than two systems of triangulation. 


Art. 43.—Final Observations on the Preceding Methods. 

Although particular forms of triangulation have been chosen 
for the various examples in the different cases of swing 
bridges, yet the conclusions reached and the principles estab- 


FINAL OBSERVATIONS ON THE PRECEDING METHODS. 22X 

lished are perfectly general. They are applicable to any form 
of triangulation, and to either the deck or through form of 
bridge; they also apply whether the two arms are of the 
same length or of unequal length, the panels being either uni¬ 
form or irregular. It is only necessary to bear in mind what 
may be called the “ local” circumstances of any given case; 
these do not, however, affect the general principles. As a 
single illustration—if the bridge is of the “ deck ” form, those 
web members which intersect in the lower chord take their 
greatest stresses together; if of the “ through ” form, those 
which intersect in the upper chord take their greatest stresses 
together. 


CHAPTER VII. 


CONTINUOUS TRUSSES OTHER THAN SWING BRIDGES. 

Art. 44.— Formulae for Ordinary Cases — Reactions—Methods of Pro¬ 
cedure. 

On account of the doubtful utility of fixed continuous 
trusses, and the extreme rarity of their occurrence in Amer¬ 
ican practice, general directions and formulae only will be 
given. It will be assumed that the moment of inertia (/) and 
coefficient of elasticity (A) are constant; it will also be as¬ 
sumed that the points of support are all in the same level, 
as it has been shown in Appendix I to what cases the result¬ 
ing formulae apply. 



16 


m 


nr 

s 


Fig. i. 

Eq. (17) of that Appendix, after introducing these condi¬ 
tions, gives, in connection with the notation of Fig. 1, the 
following equations: 


2 All (At 4) T AT 2 4 + A=o .( 1 ) 

ATj/g + 27V 2 (4 + 4) + AI§ 4 + B — o.( 2 ) 

4 + 2 T /3 (4 + 4) + MJi + C — o.( 3 ) 

M Z U + 2 T /4 (4 + 4) + Ah 4 + D— o . ( 4 ) 

Ahk 4* 2 T /5 (4 + 4) + Tf 6 4 + E— o.( 5 ) 

etc. + etc. + etc. + =0. 


The various values of M are the bending moments existing 
at the supports indicated by the subscripts; the moment at o 
is evidently nothing, since the truss is there simply supported. 

224 










FORMULAE FOR ORDINARY CASES. 


225 

The quantities A, B, C, etc., have the following values, as 
Eq. (17), of Appendix I, shows: 

A =jhp(l? - z*)z+ jhp(li - z*)z . . (6). 

4 4 

B = \hp(l? - P)g + yhp(li ~ P)g . . ( 7 ). 

4 4 

C =j2P(li-P)z + y2P{l?-z*)z . . ( 8 ). 

4 4 

D = P(/* - + ±2P(/> - P)g . . (9). 

4 4 

E =\*kP{ 1 ? - P)z + P(li - P)z . .( 10 ). 

4 4 

Etc. = etc. 4 - etc. . . 


The Eqs. (1) to (5) show, since the end moments are zero, 
that whatever the number of spans, there will always be as 
many of those equations as there are unknown bending mo¬ 
ments over the points of support. Those moments, there¬ 
fore, may always be found, and, consequently, the reactions 
which depend upon them. These reactions are the main 
objects of search. It will be necessary, then, to determine 
the bending moments at the points of support. 

From Eq. (1): 



f! __ nf 2 (4 + 4) 
4 1 4 ’ 


(11). 


By inserting this value of in Eq. (2), there at once 
results: 



2 (4 + 4) A — l 2 B 


4 4 



4 2 ~4 (4+4) (4+4) 1 

4 4 ) 





226 


CONTINUOUS TRUSSES. 


These values of M 2 and M z inserted in Eq. (3), give: 

^ _ [ - 4 2 +4 (4+4) (4 + 4)] A - 24 (4 ■+ 4) B +4 4 C 
m> ~ 777 

/ 2 ^ 3*4 

f — 8(4 4- / 2 )(/ 2 + 4 )(4+4) + 2/2(1% +4) + 2 4" (4 + 4) 1 

( 444 3 

Again, Eq. (4) gives, after inserting in it these values of 
M z and M x : 


[- iU (4 + 4) - 2 / 4 2 (4 + /s) + 8 (/a + 4) (/, 4- 4) (4 + /.)] A - [ - 44 2 + 

4 4 (4 4- 4) (4 + 4)] 4? + 2 44 (4 + 4) C + 444 

444/5 



- 16 (4 + 4) (4 4- 4) (4 + 4) (4 + 4)+44 2 (4+4) (44-4)+ 
4 4 2 (4 + 4) (4 + 4)+44 2 (4 + 4) (4 + 4) - 4 2 / 4 2 
4444 


(14). 


Any bending moment may thus be found. 

It is seen that all moments are given in terms of M lt which 
is still unknown. However, the bending moment at the other 
free end of the truss, from 0 , Fig. 1, will be zero; conse¬ 
quently its general expression, put equal to zero, will give M x 
in terms of A, B, C, etc., and the lengths of the different spans, 
i. e. } in terms of known quantities. When M t is known, all 
the other bending moments are at once given by Eqs. (11), 
(12), (13), (14), etc. 

As an illustration, if there are five spans, M 5 = o and Eq. 
(14) will at once give M x . Eqs. (n), (12) and (13) then give 
the other moments desired. 

Another method may be followed by which a less number 
of equations will suffice for a greater number of spans. For 
example, the Eqs. (11), (12), (13) and (14), with a similar 
value for M 6 are sufficient for the solution of a case of ten 
spans. 

Let A' be the quantity corresponding to A , which would 
appear in the equation involving M 9 and M Q , in a continuous 







FORMULAE FOR ORDINARY CASES. 


227 


truss of ten spans, and corresponding to Eq. (1). Let B \ C ', 
D ' represent similar quantities in equations corresponding to 
Eqs. (2), (3) and (4). The following five equations may then 
be written by the aid of Eqs. (1) to (5): 

2T/9 (/ 10 + 4 ) + M s / 9 + A = o . . (15)* 
M % k + 2J/ 8 (/ 9 + / 8 ) 4- AT? 4 + B ' = o . . (16). 
M 6 l$ + 2(/ 8 + 4) + T/g/jr -f C' — o . . (17)* 
M 7 1 ^ + 2 J/ 6 (4 + / 6 ) + M 5 / 6 + D — o . . (18). 
h + 2J/5 (/ 6 + / 5 ) + M 4 l 5 + £ = o . . (19). 

A value for M 5 may be written by changing, in Eq. (14), 
Ml to M 9 , 4 to / 10 , 4 to 4 , 4 to 4 , 4 to 4, 4 to 4 , A to A', B to 
B’, C to C', and D to D'. A value of M 6 in terms of M x 
would be equal to the value of M i} given by Eq. (13), with 
exactly the same changes made, in so far as the same quan¬ 
tities appear. These pairs of values of the two quantities M$ 
and T/ 6 , equated, would give two equations from which M l 
and M 9 could be immediately deduced. All the other mo¬ 
ments would then follow. 

The Eqs. (11), (12), (13) and (14), are sufficient in them¬ 
selves for the solution of a case of nine spans, in the manner 
just indicated. 

The preceding operations represent the most direct method 
of finding the bending moments over the points of support. 
All things considered, it is probably as short as anything that 
can be derived. 

Prof. Merriman has, however, given a more elegant method 
by the use of so-called “ Clapyronian numbers.” Any method 
involves sufficient tedium. 

The preceding formulae will be very much simplified if a 
single weight, only, rests upon some one span, since all the 
quantities, A , B , C, etc., except two, will then disappear. 

The various reactions may be immediately determined by 
Eqs. (21)—(27) of Appendix I., after the bending moments are 
found. And when the reactions are known, the stresses in 
the individual members, for a given condition of loading, are 
found precisely as for a simple truss supported at each end. 



228 


CONTINUOUS TRUSSES. 


If the ends of the truss are not simply supported, the end 
moments must be known, else the problem will be indeter¬ 
minate. In such a case the preceding methods are in no wise 
changed, but the end moments, instead of being zero, will 
appear as known quantities. 


CHAPTER VIII. 

ARCHED RIBS. 

Art. 45.—Equilibrium Polygons. 

Preliminary to the specific treatment of arched ribs it 
will be necessary, first to consider some general principles re¬ 
garding equilibrium polygons for any given system of vertical 
forces, and then those involved in the theory of flexure. 

In the figure below, let AB . K be any straight, 

simple beam, subjected to the action of the vertical forces B, 
C, D y etc. Let x be measured from any section positive and 
horizontal toward A, and let P signify any external force such 
as the reaction at A or any of the forces applied to the beam ; 



then will 2 Px represent the bending moment to which the 
beam is subjected at the section denoted by x. Now let there 
be imagined any force T acting parallel to AEK, and let the 
moments 2 Px be taken at each of the points B, C, Dy E, E, 
G, H. Then if the quotients of those moments divided by 
the horizontal component of T be supposed represented by 
the vertical lines bB, cC, dDy etc., respectively, will the poly¬ 
gon AbcdefghK be one equilibrium polygon for the given sys¬ 
tem of loads ; so that if the beam AK were displaced by a tie 
in which exists the stress T, and the given loads hung from 
the joints b, c, dy etc., the whole system would be in equilit> 
rium. 

229 








230 


ARCHED RIBS. 


In order to establish this, it is only necessary to show that 
no piece of the polygon is subjected to bending; for if that 
is the case, the line of action of the resultant stress must co¬ 
incide with its centre line. 

Consider any portion of the system, as that lying on the left 
of the vertical line dD. Those forces which have moments 
about the point d are the external forces to the left oidD and 
the stress T in the tie AK ; the latter has a lever-arm n, equal 
to the normal distance from d to AK, and its moment is op¬ 
posite in sign to 2 Px. Consequently the resultant moment 
about d will be M — 2 Px — Tn . But by construction 2 Px — 
Tn, hence M — o; and the same is, of course, true of every 
other joint. If T h is the horizontal component of T, then 
evidently Tn = T h {dD) — TZPx. 

If v be the general representative of the vertical ordinates 
bB , cC, etc., then, in general, 


2 Px 



but T h is a constant quantity. From these considerations 
follows this important principle : 

The vertical ordinates of the equilibrium polygon of any sys¬ 
tem of vertical loads are proportional to, and may represent, the 
bending moments found at the various sections of a beam sub¬ 
jected to the action of the same system of loads , and having the 
same span. 

Since the stress T was taken arbitrarily, it is evident that 
there may be an indefinite number of equilibrium polygons 
for any given system of loads; the principle stated above, 
however, is perfectly general, and is true for all. 

Since vT h — 2 Px = constant for any given section, it fol¬ 
lows that any variation of T, and therefore T h , produces an 
opposite kind of variation in v. Hence the height of an equi¬ 
librium polygon is proportioned to the reciprocal of T or T h . 

The method of constructing the equilibrium polygon given 
above is not the most convenient, nor the one commonly 
used. The method ordinarily used is the usual one for con- 



EQUILIBRIUM POLYGONS. 


231 

structing the equivalent polygonal frame, and is the follow¬ 
ing: 

Let AK, Fig. 2, represent any span, inclined in this case 
but ordinarily horizontal, and 1, 2, 3, 4, etc., the vertical loads 
acting along their respective lines of action. In Fig. 3 let the 
portions I, 2, 3, 4, 5, 6, 7, 8, and 9 of the vertical line 1-9 rep¬ 
resent those loads taken by any assumed scale. Since BC 


represents the sum of all the applied loads, it is also equal to 
the sum of the two reactions or shearing stresses at A and K. 
In the case of the simple beam taken, those quantities will of 
course be determined by the law of the lever only. 

Suppose A'C and A'B to represent the shearing stresses or 
reactions at A and K respectively. Then draw A'P parallel 
to AK, and on it take 
any point P. From P 
draw the radial lines 

a, b, c, d, ./, as 

shown, and starting 
from A ox K m Fig. 2, 
draw the lines a, b, c, 

d, ./, parallel 

to the lines denoted 
by the same letters in 
Fig. 3. Then will Fig. 

2 represent the equi¬ 
librium polygon for the 
given span and system 
of loading. 

The line AK or PA' 
is called the closing line of the polygon. The reaction at A is 




















ARCHED RIBS. 


232 

evidently composed of the numerical sum of the vertical com¬ 
ponents in / and AK, while that at K is equal to the numer¬ 
ical difference of the vertical components in a and AK. 

The point P, from which the radial lines are drawn, is called 
the pole , and the normal distance from the pole to the load 
line BC f the pole distance. The pole distance evidently rep¬ 
resents the horizontal component of stress common to all the 
members of the polygon. 

In order that the equilibrium polygon, constructed accord¬ 
ing to the principles given above, shall exactly fit the span, it 
is only necessary that a proper observance be paid to the 
scales used. 

y? Px 

From the equations = ———it is seen that the scale for the 

■l h 

forces does not affect the height of any joint of the polygon; 
it depends only on the scale according to which x or the hor¬ 
izontal span is drawn. 

Let the line PP' be drawn parallel to BC, and let P r be 
the pole of a new equilibrium polygon ; the pole distance 
will, of course, remain the same as before. But the pole dis¬ 
tance represents the horizontal component of the stress in the 
closing line, and it has already been shown that if T h remains 



the same, v cannot vary. Hence, any movement of the pole 
parallel to the load line does not change the vertical dimensions 
















EQUILIBRIUM POLYGONS. 


233 


of the equilibrium polygon. But if the pole distance is changed , 
the vertical dimensions are changed in the inverse ratio. 

The determination of the deflection polygon of an arched 
rib with ends fixed, involves the use of an equilibrium poly¬ 
gon, similar to that required for a 
system of forces whose resultant is 
a couple. Its method of construc¬ 
tion is not at all different from 
that just given. 

In Fig. 4, let the forces I, 2, 3, 

4* 5^ 6, 7, 8, and 9, act vertically, 

BC being horizontal, and let the 
sum of 1, 2, 3, 8, and 9 be numeri¬ 
cally equal to the sum of 4, 5, 6, 
and 7. The double line, DE , in 
Fig. 5, represents the forces shown 
in Fig. 4. 

In Fig. 5, draw the line a in a 
horizontal direction through the 
upper extremity of force 1, and 
take any point on it for the pole P. From /Mraw the radial 
lines in the usual manner as shown. 

From C, in Fig. 4, draw b' parallel to b in Fig. 5, until it in¬ 
tersects the line of action of force 2. Then draw the other 
lines, c, d \ etc., parallel to c, d, etc., until the lines of action 
of the other forces are intersected, b 1 , c, ... . Ji, k\ will then 
be the equilibrium polygon for the system of forces assumed. 

It is seen that the polygon does not close. This simply 
shows that the resultant of the system is a couple, whose mo¬ 
ment is the force a, in Fig. 5, multiplied by AB (vertical) in 
Fig. 4. 

The following general principle then results : The equilib¬ 
rium polygon for any system of parallel forces whose resultant 
is a couple, is not a closed one. 

This principle is indeed true for any system of forces. 

P might have been taken at any other point, as P' in 
OP. In that case, however, the equal forces a, acting at A 
and C, Fig. 4, would be parallel to a line drawn from the 



Fig. 5. 






234 


ARCHED RIBS. 


upper extremity of force i to P'. The vertical dimensions of 
the polygon, measured from either of the forces a (in gen¬ 
eral inclined), will always be the same if the pole remains in 
the line PP\ 


Art. 46.—Bending Moments. 

An arched rib is any truss curved in a vertical plane, both 
of whose chords are convex or concave in the same direction, 
neither being horizontal; the ends may be fixed or free. 

In Fig. 2 of Art. 45, let ADEK represent an arched rib 
sustaining the loads 1, 2, 3, 4, ... . 9. Now it has already 
been seen that, so far as equilibrium is concerned, any given 
system of loading may be sustained by any one of a set of 
equilibrium polygons consisting of an indefinite number. On 
the other hand, it is evident that no polygon or arched rib 
can be drawn, which is not an equilibrium polygon for some 
system of vertical loading; but if that arched rib sustains 
some other system of loads than that which, it may be said, 
properly belongs to it, and if its,joints be prevented from 
turning, it will be subjected to bending, which will vary from 
one section to another. 

The arched rib ADEK sustains a system of loading for 
which AfK is the equilibrium polygon, hence the former will 
be subjected to varying degrees of bending at various sec¬ 
tions. When the rib ADK is subjected to the action of its 
load, stresses are developed in its different parts, whose hori¬ 
zontal components are all the same because the load is wholly 
vertical. Now if an equilibrium polygon can be found in 
which the horizontal component of stress T h is the same as 
that developed in ADEK , then all the circumstances of stress 
and bending in the latter can be determined, as will be seen 
hereafter. 

Suppose AfK to be that polygon, then let v f denote the 
portion of a vertical line intercepted between it and the arched 
rib, as DD . The moment about any point D will then be 

M= 2 Px- T h (v + v). 

But since AfK is the equilibrium polygon, 2 Px — T h v — o. 


BENDING MOMENTS. 


2 35 

.*. M = — T h v'. When the polygon lies above the rib, v r is 
negative, and, hence, M positive. 

Let the polygon which has the same value of T h as the 
arched rib be called the true equilibrium polygon ; then, since 
T h is a constant quantity for the same rib, there is established 
the following important principle: 

The bending moments to which the different parts of an arched 
mb are subjected are proportional to , and may be represented by, 
the vertical intercepts included between the rib and the true 
equilibrium polygon. 

This principle has been demonstrated for a beam with free 
ends only, but it is true also for a beam with fixed ends, as 
will now be shown. 

In order to fix the end of the rib it is only necessary to im¬ 
press upon the rib at A, the point of fixedness, the proper 
couple whose moment is m ; in the fixed rib, as in the free, 
let T h represent the horizontal thrust. The true equilibrium 
polygon for the fixed rib will be that found by increasing the 
vertical dimensions of a polygon for a free-end rib, formed by 
using T h and reactions for fixed ends, by a constant amount 

ill 

equal to . Again, taking moments about any point of 
h h 

the centre line of the rib, there will result 

m 

M= 2 Px + m - T h (v+ -=- + v) = - 7 >', 

* h 

as before. This shows that the principle stated above is true 
for both fixed and free-end ribs. In truth this equation might 
have been written first, and the special case of the free-end 
rib deduced by making m — o. 

An arched rib, then, when subjected to the action of a load, 
suffers bending in the same manner as a straight beam, but to 
a different degree. 

In so far as it plays the part of a beam, it must be governed 
by the general laws of bending or flexure. The formulae to 
be given in this connection are those approximate ones based 
on the common theory of flexure, and found in the ordinary 
works on that subject. 



236 


ARCHED RIBS. 


Art. 47.—General Formulae. 

Let 5 denote the total shearing stress at any section, P any 
applied load or external force, M the bending moment at any 
section, and D the deflection found above; then the six gen¬ 
eral equations of flexure demonstrated in the Appendix on 
the Theorem of Three Moments, some of which are made use 
of in the graphical treatment of arched ribs, are the following: 


5 

M 


P' = 


= 2P, 

= 2 Px f 
M 


El y 


2nP ' = 2 


nM 

~EJ' 


D 


= 2nP x = 2 


nMx 

~eT 


D h =2 nPy. 

If the beam is originally straight and parallel to the axis of 
x y n becomes dx and y — o. 

As usual, E and I represent the coefficient of elasticity and 
moment of inertia of the cross-section, respectively. 

The limits of the summations are the section considered 
and any section of reference. The quantities S, M, P’ and 
D then refer simply to that portion of the beam over which 
the summation extends. 

One very important deduction is to be drawn from the above 
equations, or rather from the second and fifth of them. It is 

n M 

seen from these two equations that -gj- stands in the same 

n M 

relation to D that P does to M. Consequently if —gj be put 







ARCHED RIB WITH ENDS FIXED. 


23 7 


in the place of P in any graphical construction, D will be 
represented in the place of M. If\ therefore , an “ equilibrium 

yy flM 

polygon ” be constructed for any span by taking as loads 

instead of P, the vertical ordinates of the polygon will repre¬ 
sent the deflections at the sections denoted by the corresponding 
values x. 

This polygon may be called the “ deflection polygon,” and 
its construction plays a very important part in the determina¬ 
tion of the true equilibrium polygon for an arched rib, in the 
majority of cases. 


Art. 48.—Arched Rib with Ends Fixed. 


The ends of an arched rib or any girder are considered 
fixed when the angles formed by their centre lines with any 
assumed line, at the fixed sections, do not vary under any 
applied load. 

Let BAD , PI. V., be the centre line of any arched rib ; it 
will, of course, be considered fixed at the points B and D. 
This line may be any curve, though for convenience the arc 
of a circle has been drawn. In the demonstration no atten¬ 
tion whatever has been given to the character of the curve, 
so that it is equally applicable to any other curve. 

In the present case the centre line of the rib will be divided 
into equal parts for the application of the load, and each of 
those parts will be n ; consequently that quantity will have a 
finite value, and the results will not be strictly accurate, 
though near enough for all technical purposes. 

The piers or points of fixedness are supposed to be immov¬ 
able, whatever may be the character of the load ; but if that 
is the case, the summation of the strains at any given dis¬ 
tance from the neutral axis of the rib, considered as a truss, 
taken throughout the whole length of the rib BAD, must be 
equal to zero. Take that distance as unity, then there re¬ 
sults, for one condition, since n is constant, the equation: 



238 


ARCHED RIBS. 



B 




the moment of inertia of the cross-section of the rib is supposed 
to be the same throughout its entire length. More will be said 
on this point hereafter. 

It has already been shown that the area included between 
the equilibrium polygon and the curve BAD is made up of 
vertical strips, whose lengths (the vertical intercepts) repre¬ 
sent the actual bending moments at the different sections of 
the rib. Hence represents the sum of those vertical 

lengths or intercepts drawn at the points to which the moments 

M belong , and the equation M — o shows that the sum on 

one side of BAD must be equal to that on the other. 

But, as will be seen, there may be an indefinite number of 
equilibrium polygons which will fulfill this condition ; conse¬ 
quently at least one other condition must be obtained. 

Since the points B and D are fixed, the sum of all the de¬ 
flections, both horizontal and vertical, taken between those 
two points, must be equal to zero. It has been shown that 
the vertical deflection at any point, when n and I are con- 

71 

sidered constant, is D — ~pgp ^Mx ; also, from the reasoning 

applied to the curved girder, that the horizontal deflection is 
n 

D h == —j 2 My. Now, when these summations extend from 

B to D , since those points are fixed, both D and D h must equal 
zero. The three equations of condition, then, which must be 
fulfilled for the rib, are: 




2 P n My= o. 


It has already been stated, and it is evident without much 




ARCHED RIB WITH ENDS FIXED. 


239 


thought, that any polygon whatever is an equilibrium polygon 
for some load. 

Hence consider BAD , PL V., an equilibrium polygon for its 
proper load, and consider it subjected to that load ; denote its 
moments by M h . 

Again, suppose the polygon a , a , a \ etc., to be the true 
equilibrium polygon for the given load, and denote its mo¬ 
ments, represented by the vertical ordinates drawn from its 
closing line, by M a . 

Then, from the principle which precedes the equation M = 
— T h v' in the general discussion of equilibrium polygons, 
there follow the equations: 


M = M a — M b , 

Mx — M a x — M b x . 

/. 2 b d Mx = 2 “ v M a x - 2 D o M b x = o. 

Or, 2 B D M a x = 2 *M b x. 

In the same manner, 2 ^M a y = 2 B p M h y. This last equation 

will be used in fixing the pole distance of the true equilibrium 
polygon. 

It must be remembered that M represents the actual mo¬ 
ment to which the rib is subjected at any point. 

The application of these two conditions will be shown in 
the course of the construction of the true equilibrium poly¬ 
gon, as they are needed. 

In the figure of PI. V., let the scale for linear measurements 
be 10 feet to the inch, and the force scale 15 tons per inch. 
The curved centre line of the rib is divided into ten equal 
parts of 10.95 feet each, and that is the constant value of n. 
The load is not therefore uniformly distributed. The panel 
length, or horizontal distance between the points of appli¬ 
cation of the loads, is thus a variable quantity. If the versed 
sine of the centre line of the rib is small, n may be taken 
equal to the span divided by the number of panels. But if 
the versed sine is not larger, even, than in the present case, n 


240 


ARCHED RIBS. 


cannot be so taken without sensible error, as will be seen. 
The other data are as follows: 

Span = ioo feet. 

Radius = 75 feet. 

Angular length of curve = 83° 37'. 

Panel fixed load = 4 tons. 

Panel moving load = 10 tons. 

Centre rise of rib = 19.1 feet. 

In the figure BD is the span, and C the centre ; b', b", b"', 
etc., are the panel points equidistant in the curve, and through 
which the loads are supposed to be applied. 

Now the actual moment area for the arched rib is supposed, 
really, to be the difference between the moment area of the 
true equilibrium polygon for the applied loads and that of the 
rib itself considered as an equilibrium polygon for the proper 
load ; both systems of loading being supposed applied to a 
straight beam fixed at each end. 

The first portion of the problem which presents itself, then, 
is to determine the true equilibrium polygon for the given 
load. The construction will first be made, and it will then be 
shown that the two conditions given above are satisfied. 

Let the moving load cover the left half, BC, of the span, 
and suppose the half panel loads at B and D to rest directly 
on the abutments. According to the scale taken, lay off B 6 
equal to 7 tons, half the total load on the panel Ab lv , and 
B 5 equal to half the fixed panel load on Ab Yl ; then lay off 
6— 10, divided into four equal parts, equal to the four equal 
panel loads on b 1 , b", b"\ and b u . In the same manner lay 
off 5 — 1, equal to the four fixed panel loads on the right half 
of the span. Assume C as a convenient pole, and draw the 

radial lines from it to 1, 2, 3, 4,.10. Starting from C , 

draw C — 6 until it intersects a vertical through b iY at a 4 ; from 
the latter point, a 4 a 3 parallel to C 7 until it intersects the ver¬ 
tical through b ’"; proceed in the same manner until the poly¬ 
gon ECF is drawn. If the ends were free EF would be the 
closing line, but one must now be found that will satisfy the 



ARCHED RIB WITH ENDS FIXED. 


241 


condition 2 M a — o, or, in other words, the sum of the verti¬ 
cal intercepts drawn from the closing line downwards must 
be just equal to the sum of those drawn from the same line 
upwards. 

The proper closing line is easily located by trial. If 
Cc y — 0.5 inch and vv' = 0.96 inch, there results: 

2 B D M a = l vv' + - a./' - a,c'" - a t c‘ y - Cc" - atf* - a 

+ + ayc lx 4- £v"v'” = 1.88 — 1.89= — 0.01 inch. 

This sum is sufficiently near zero. 

The lines vv' and v"v" are drawn vertically through points 

71 

b and b x , distant — from B and D on the curve BAD , and 

4 

their halves are taken because in the summation 2 B n M a there 

D a 

71 71 UTJ* 

would appear terms — x vv' and — x v"v'" f or n x — and 

2 2 2 

v"v"' 

n x-. 

2 

Similar terms will hereafter appear in similar summa¬ 
tions. 

The closing line HK then satisfies the condition D M a - o 
for the equilibrium polygon ECF. 

There still remains the condition 2 B D M a x = 2 B D M b x . This 

equation will be satisfied by making each of its members 
equal to zero. The closing line HK must, then, also make 

2 D M a x = o. This simply means that the vertical ordinates 

of the polygon a , measured from HK , multiplied by their 
horizontal distance, from D , will form a sum equal to zero 
when their products are added. If the ordinates below HK 
are taken positive, as v"v", a 9 c ix , a x c \ etc., and those above, 
as negative, and if the ordinates and distances be taken 
by scale from the drawing, there will result, nearly, 

v'v ,n x De + x De' + a^c' x De y + etc. = + 99.7 

a 7^ x De" + x De " + etc. 

16 


= — 102.0 




242 


ARCHED RIBS. 


The numerical values are nearly enough equal, and the line 
HK will be taken as the proper closing line. 

The next step is to find the closing line for the curve BAD 
of the rib, considered as an equilibrium polygon, which will 
satisfy the same general conditions. 

Using precisely the same method of procedure as for the 
polygon ECF, the line H"K" is found to be the one desired, 
for that line makes the sum of the intercepts above it just 
equal to the sum of those below it. 

Ac 5 is about 0.62 of an inch. 

For this curve the summation may be written : 

2 B D M b = bH' + 2 b'c x + 2 b"ct - 2 c % b'" - 2 cJF - Ac % = o. 


Since the curve is symmetrical in reference to A, the static 
moments of the ordinates on one side of H"K", about DK", 
will evidently be equal to the same moment of those on the 
other side. 

The second condition may now be applied. That condition 

D D 

is 2 D M a y — 'E D M b y. It has already been shown that if M a 

and M b be considered loads applied at distances y from the 
assumed origin, the ordinates of the equilibrium polygon so 
constructed will represent the quantities EM a y = D a , or 
EM b y = D b . 

Through A, therefore, draw the horizontal line RAS. As¬ 
sume any line, as AC, as the closing line of the deflection 
polygon, and lay off At equal to a half of vv. Also make tt 


equal ca x ; t't" equal c"a 2 ; t"t equal a s c etc. t"t"’ is 
measured in a direction opposite to that of the preceding, 
because it represents a moment of an opposite sign. In the 
same way At measured to the right of A is equal to a half of 
v"v f "; tt x equal to ; t t t n equal to a%c ynx , etc.* Draw 
bb x and note its intersection C" with AC. From C draw 
C"dl parallel to Ct until it intersects a horizontal drawn 
through b' in d' \ draw d/d” parallel to Ct' until it intersects 
a horizontal through b" in d ”; draw d'/d’" parallel to Ct” 


* If the drawing had been made accurately, / iv 7i V would have been exactly 
equal to Cc v . 



ARCHED RIB WITH ENDS FIXED . 


243 


until it intersects a horizontal through b'", etc. The polygon 
C'd'd etc., will intersect the horizontal RAS at a point 
distant from A, on the left of it, from which / is drawn to a 
point on the right of A at the intersection of the deflection 
polygon formed by using /, t n t u , t m , as before, and which is 
shown in the figure. The sides of the deflection polygon on 
the right of CA are parallel to radial lines drawn from C to 
the points t, t t , t u , t ln . The distance (/ — /„) represents, in an 
exaggerated manner, the horizontal deflection of the end of a 
vertical beam fixed at C, whose length is CA , and which is 
subjected to the bending moments vv, c'a t , c"a 2 , etc., at verti¬ 
cal distances from C equal to the heights of b, b', b", etc., above 
BD. In the same manner, l lt represents the same quantity 
for the same beam when subjected to the corresponding mo ¬ 
ments on the right-hand side of A C. 

The line /, can be determined with less work and more sim¬ 
ply when the meaning of the construction is once clearly 
seen, by laying off, on the left of A, as before, loads repre¬ 
sented by the algebraic sums (At + At), ( tt' + tt), (t't" + t f t n ), 
etc., and then drawing the equilibrium polygon as usual. The 
distance from A to the intersection of the polygon with AR 
will then be equal to l r 

The deflection polygon d'd n d"'d iy is constructed in pre¬ 
cisely the same manner as the preceding. Make As equal to 
a half of bH '; ss equal to cb 1 ; s's" equal to c 2 b" ; s"s" equal 
to c 3 b r ", etc.;* then draw radial lines from those points of 
division to C. The point d’ is at the intersection of C"d 
drawn parallel to Cs , with a horizontal line drawn through b ’; 
d'd" is drawn parallel to Cs' until it intersects a horizontal 
line drawn through b", and the other sides of the polygon are 
constructed in the same way. 

The polygon cuts the horizontal line RAS in a point distant 

from A. There will, of course, be another deflection poly¬ 
gon, precisely the same as the last, on the right-hand side of 
AC, found by taking )K"D, r 9 £ ix , etc., and laying them off 
from A towards 5. 


* With a sufficient accuracy of construction, As ly would equal a half of Ac 5 , 



244 


ARCHED RIBS. 


If the intercepts used in the deflection polygons for M a and 
M b , represent those moments by the proper scale, then by the 
same scale CA will not, in general, represent the true pole 
distance. But this fact has the same proportional effect on both 
/ and /. Consequently any result depending on the equality 
of / and l t will not be affected. 

Instead of using Ac 5 and Cc y in the manner shown, greater 
accuracy might have been attained by taking half intercepts 

n 

at the distance — on both sides of A and C. Such an oper- 

4 

ation, however, is unnecessary in all ordinary cases, since 
moments in the vicinity of A and C have very little effect on 
the horizontal dimensions of the deflection polygon. Mo¬ 
ments, on the contrary, in the vicinity of HH ", have great 
effect. 

i 

B B 

Now l t represents 2 M a y ; and /, 2 M b y ; and in order 

that the second condition may be satisfied they should be 
equal. Since /, is less than /, it shows that the quantities M a 
are too small, or, in other words, the pole distance BC is too 
large. This last statement is evidently true, if it be remem¬ 
bered that the pole distance is inversely proportional to the 
vertical ordinates which represent the moments. 

Lay off, therefore, on AC produced, the distance CM equal 
to l t , and draw through M the horizontal line MN. With a 
radius CN equal to /, draw the arc of a circle cutting MN in 
N, then produce the line CP. All moments represented in 
the lower equilibrium polygon a will have to be increased in 
the ratio of CN to CM. To make this reduction, draw a 
horizontal line, for instance, through c y until it cuts CN in/, 
then make c* > a y equal to Cp ; a y will be one point in the true 
equilibrium polygon. All other points might be found in the 
same way, but having found one point, as a y , a much shorter 
method may be used. 

It has already been shown that the vertical dimensions of 
two equilibrium polygons for the same loading and span are 
inversely proportional to the pole distances, the vertical 
dimensions being measured from the closing lines. In the 


ARCHED RIB WITH ENDS FIXED. 


245 


figure the vertical dimensions of the polygon ECF must be 
increased in the ratio of l x to /, or in the ratio of Cc y to Cp. 
Hence, on CN produced make equal to BC, and draw the 
horizontal line OP cutting CM produced in 0 , then will CO 
be the pole distance for the true equilibrium polygon. 

In order to find the true pole, make CL parallel to HK , 
then draw LC' parallel to BC* and equal to CO; the point 
C' will be the true pole. 

According to previous principles, the reactions or vertical 
shearing stresses at B and D will be L — 10 and L — 1 respect¬ 
ively, and, since the closing line must be parallel to BD in the 
true equilibrium polygon, LC' must be parallel to BC. From 

C' draw the radial lines shown to 1, 2, 3,.and 10; 

these lines will be parallel to the sides of the true polygon ; 
i. e., draw a y d y parallel to C' 6 until it cuts a vertical through 
b iy ; a iy d" parallel to C '7 until it cuts a vertical line drawn 
through b'" ; a y aA parallel to C' 5, etc. The polygon acid' 

. a xx a x so formed will be the true equilibrium 

polygon. 

As a check some of the points as d or a" should also be 
found by the previous method. Thus make Cg equal to c'a t , 
and draw gp' parallel to CD ; Cp ' should then be equal to ca\ 
Other points may be treated in the same manner. 

It may now be seen that the polygon add' . a x 

satisfies the three conditions 2 M= o, 2 d Mx — o, and 

p 

2 B My — O. The first conditions are evidently satisfied by 
the method of locating the closing lines in the lower polygon 
ayi 2 . a 9y and the curve BAD, for any intercept be¬ 

tween the curve and the upper polygon a is the difference 
between two intercepts each of which belongs to a sum equal 
to zero, hence the sum of those intercepts is zero. The second 
condition is satisfied by the location of the point C'. 

Another check on the degree of accuracy attained in the 
construction is found in what has just been said, i. e., the 


* The reason for the true closing line being horizontal is given, though for 
another purpose, on page 273. 







246 


ARCHED RIBS. 


moment area lying above the curve BAD must be just equal 
to that lying below it. 

By actual measurement LC' is equal to 3.65 inches, hence 
the constant horizontal component of stress in any portion of 
the rib is 54.8 tons. The resultant stress at any point of the 
arched rib is equal to 54.8 tons multiplied by the secant of 
the inclinations at that point. The bending moment at any 
point to which the rib is subjected is found by multiplying 
the vertical intercept between the equilibrium polygon and 
the curve by (54.8 tons = T h ). For example, the actual mo¬ 
ment to which the rib is subjected at U is (b d x 54.8). 

The line of action of T h (54.8 tons in this case) is, of course, 
along the true closing line H"K ". This is an important 
matter, as will hereafter be seen. 

* 

A line drawn through C parallel to EF would cut off on 
the load line the reactions which would exist were the ends 
free. 

The reaction at B, in the present case, is thus seen to be 
much greater than would be found in the case of free ends. 

A point in the vertical line passing through the centre of 
gravity of the load is found at the intersection of the sides a 
a and a ix a x in G. G\ at the intersection of Ea x , and Fa 9 , 
prolonged, is in the same vertical line. 

The pole of the deflection polygons might have been at A, 
and the moments laid off from C, in which case the half of 
Ac 5 , and the same of Cc y , would have been the moment dis¬ 
tances adjacent to C. Precisely the same value for LC' would 
probably not be found, because the sum 2 My is not con¬ 
tinuous, and consequently not exact. For this reason it would 
be better in an actual case to divide the real panel lengths 
into two or more equal parts in order to find the true pole 
distance, LC', and consequently the true equilibrium polygon. 
T h can then be used to find the stresses in the members of 
the actual rib in a manner that will hereafter be shown. 

The diagram should of course be drawn to as large a scale 
as possible, and it may often be advisable to exaggerate the 
vertical scale so as to make the intersections of the true poly¬ 
gon and curve well defined. 


ARCHED RIB WITH FREE ENDS. 


24 7 


The effect of such an exaggeration may easily be shown 
since the various steps of the construction remain precisely 
the same. Suppose A C to be in times as large as it would be 
made by the scale according to which the span BD is laid off; 
m denotes the degree of exaggeration. The distance / will be 
m times as great as it ought to be, and consequently the 
height of the equilibrium polygon will be increased beyond 
its true value in the same ratio. But if the true height is 

only — of that found, the true pole distance will be m x CO , 
m 

and LC ! must be made equal to that. 

The span has been supposed half covered by the moving 
load, but any other portion might have been taken as well. 
It is to be noticed that the method is perfectly general, and 
entirely independent of the character of the curve BAD , or 
of the loading. 


Art. 49.—Arched Rib with Free Ends. 


The treatment of the arched rib with free ends is not dif¬ 
ferent in any respect, except one, from that given in the pre¬ 
vious case. The exception is this, that the condition 2 M = o 
must be omitted, since the bending moments at the free ends 
must disappear. 

In this case, again, the centre line is divided into equal 
parts. Observations made under this head in the preceding 
Article apply here also. 


The expression ZznP' — ^ 


nM 
~EI ’ 


sometimes called the bend¬ 


ing, is of course based on the common theory of flexure, and 
denotes simply the difference of inclination of the neutral 
surface of a straight beam at the two sections indicated by 
the limits of the summation; with a constant moment of 


inertia it is usually written 



The condition of fix¬ 


edness of the two ends of the ribs requires the position of the 
neutral surface to remain unchanged at those two sections, 
consequently ' 2 nP' must be equal to zero between those 





248 


ARCHED RIBS. 


limits. In the case of free ends, the position of the neutral 
surface may be any whatever consistent with the elastic prop¬ 
erties of the material at those sections. The middle points 
of the free-end sections must, however, retain their primitive 
positions, or the summation of the deflections, either horizon¬ 
tal or vertical, between those points must be equal to zero. 

The only remaining conditions, therefore, are 'E D D nMy = o = 
2 *nMx. 

But the ends of the rib may have any relative vertical 
movements, whatever, without changing the circumstances of 
bending. Consequently the only condition to be fulfilled is, 

nMy — o. 

The same amount and proportion of loading will be taken 
as in the previous case; the same radius, span, notation, and 
scale will also be taken. The figure of PI. VI. represents the 
construction. The moving load is assumed to cover the half 
span BC. 

As before, take C as the pole for the trial polygon, then 
make B 5 equal to a half panel fixed load, supposed applied 
at A, and B 6 a half panel (fixed + moving) load, supposed 
applied at the same point ; also make 5—4 equal to load at 
b Y \ and 6 — 7 the load at b iY , etc. 

Draw radial lines from C to the points of division, 1, 2, 3, 
4, 5 • • • • 10, and construct the polygon E, a 1} a 2 , a 3 , . . . . 
F y precisely as before ; in truth it is exactly the same polygon 
that was used in the preceding case. Since there can be no 
bending moments at B and D, the equilibrium polygon must 
pass through those points, hence EF is the closing line of the 
trial polygon, ECF. 

As has been seen, the only condition to which the equilib- 
rium polygon is subject is 2 ^ nMy = o ; or, as before, as n is 
constant. 

2 *My = 2 B D M a y-Z*M>y = o. 

or, K M °y= 

The method of constructing the deflection polygon is pre- 


ARCHED RIB WITH FREE ENDS. 


249 


cisely the same as that followed in the previous* case, only in 
the present one a half of the ordinates representing M a and 
M b will be laid off from A in order to keep all the points s 
and t within the limits of the diagram. 

As the half intercepts at the distance — from B and E are 

4 

very small, and as their omission will lead to simplicity in the 
diagram, and not cause much of an error, they will be neg¬ 
lected. In an actual case, however, the omission should be 
made with caution. 

Since the moments at B and D are zero, make As equal to 
\c<Jb', ss’ equal to \c % b" , s' s' equal to \c z b"' , s"s"' equal to \cjd y ; 
then draw horizontal lines through b'b", b"b"‘ , etc., cutting 
AC as before. As the end moments are zero, d‘ will be on 
AC y then d'd" will be parallel to Cs, d"d'" will be parallel to 
Cs'y and so on until the point on the line RS is reached. 

The two portions of the deflection polygons for the mo¬ 
ments M a will not be similar, yet it is only necessary to con¬ 
struct a single deflection polygon, as was shown in the pre- 

ceding Article. The sums 2 D M a y and 2 D M b y may each be 

divided into pairs of terms, each member of the pair having 
the same value of y ; this may be done for any case in which 
the moments may be taken in pairs. The moments in each 
pair of terms will of course be located equidistant from AC. 
Make, therefore, At equal to \ {a x c' + # 9 <: ix ), tt t — to \ ( a. 2 c" + 
# 8 c v111 ), t,t n equal to J a z c‘" + tf 7 r vU ), and t n t m equal to J {a x d y 4- 
a 6 c yi ). Draw radial lines as shown, and make d t ‘ dj' parallel 
to Cty d'/dj" parallel to Ct ,, etc., until the point d, y is reached. 

Lay off on AC produced, CM equal to Ad, y , draw MN par¬ 
allel to BDy and with a radius CN equal to 2 (Ad y ) find the 
point Ny and produce CN through that point. In order to 
find the true equilibrium polygon it is only necessary to in¬ 
crease the ordinates of the trial polygon in the ratio of CM 
to CN. 

Hence draw c y p parallel to CD , then make Ca y equal to Cp\ 
a y will be one point in the true equilibrium polygon. In the 
same manner make Cg equal a z d"y and determine p' as before, 


ARCHED RIBS. 


2 50 

then make cgi"' equal to Cf ; a" will be another point in the 
true polygon. A shorter way to proceed, however, is the one 
indicated in the previous case. Draw CL parallel to EF, then 
a line parallel to BC through L . Make CP equal to BC , and 
draw OP parallel to BD ; OC is the true pole distance. Make, 
therefore, LC' equal to OC , and C' will be the true pole from 
which radial lines are to be drawn to 1, 2, 3, 4, 5, . . . . 10. 

Starting from any given point, as B } make Ba' parallel to 
C' 10, a'a" parallel to C' 9, etc., etc.; the side parallel to C' 1 
should pass through the point D. 

The two methods should be used to check each other, as 
was indicated in the previous case. 

Since LC' is 3^ inches, T h — 3J x 15 = 52.5 tons, and its 
line of action is evidently BD. 

All those directions of a general character which accompany 
and follow the construction in the preceding case apply with 
equal force to the present one and those which follow. It is 
particularly important in the graphical treatment of all arched 
ribs to make the polygons approach as nearly their ultimate 
limits, i. e. y curves, as possible ; for that reason it will be 
advisable in most cases to divide the actual panels into two 
or more equal parts in the search for the true equilibrium 
polygon. 

Art. 50.—Thermal Stresses in the Arched Rib with Ends Fixed.* 

Thermal stresses are those stresses which are co-existent 
with any variation of temperature, in the structure consid- 

* In reality the deflection which produces stress, in the case of variation of 
temperature, is not the whole deflection. If the points B and D, in PI. VII., 
were free to move, there would be no thermal stresses, but there would be deflec¬ 
tion. This deflection, in the present case, would be (if 100 units become 100.12204 
for an increase of iSo° F.) 

/ 165 \ 

(100 + —g— 0.12204 ) 100 x 19.1 — 19.1 = .021 ft. 

Strictly speaking, the deflection to be used, then, for a change of 165° F. is 
0.25 — 0.021 = 0.23 ft. The difference is so small, however, that it may be 
neglected, especially since the error is a small one on the side of safety. These 
observations are general and apply to all cases. 



THERMAL STRESSES IN ARCHED RIB . 


251 


ered, and whose values depend upon that variation. Any 
variation of temperature in the material of which an arched 
rib is composed will cause a variation in its length, and con¬ 
sequently a deflection at any given point. Although the 
temperature is supposed to change, yet the ends of the rib 
are supposed to remain in their normal positions, so that the 
general conditions IznM — o and 2 nMy = o — 2 nMx, hold 
for thermal stresses ‘as well as for any other, remembering 
that n is constant. 

Any change of form, such as that arising from the applica¬ 
tion of loading, will cause extra stresses, which are to be de¬ 
termined in precisely the same manner as that used for ther¬ 
mal stresses ; in fact, they belong to the same class of stresses. 

Every kind of material has its own coefficient of linear ex¬ 
pansion ; wrought iron, for instance, expanding .12204 units 
in every 100 for a change of temperature from 32 0 to 212° F., 
while tempered steel gives the empirical quantity. 12396 for the 
same conditions. (D. K. Clark, “ Rules, Tables, and Data.”) 

The arched rib shown in PI. V., and already treated for 
ordinary stresses, will be supposed to be of such a material 
and to be subjected to such a change of temperature that the 
point A will suffer a vertical deflection of 3 inches. It is a 
matter of indifference in which direction the deflection takes 
place; it will be supposed upward in the present case. 

The effect of the thermal variation is to cause bending 
moments at the various sections of the rib from which the 
deflection results. Also, to keep'the ends in their original 
positions requires the existence of a horizontal force, such as 
the stress which may be supposed to exist in a horizontal tie. 
The stresses and bending, then, caused by thermal variations 
are the same as those which would be caused by a horizontal 
force having the proper line of action ; the problem then 
resolves itself into finding the proper value and line of action 
of this horizontal force. 

The figure to be used, and which will be referred to, is that 
of PI. VII., and represents the same rib precisely as PI. V. 
For the sake of greater accuracy, n will be taken half as great 
as in Pis. V. and VI.; its value will then be 5.47 feet. 


252 


ARCHED RIBS. 


It will first be assumed that the points b, b ly b 2 , etc., are at 
a uniform horizontal distance apart of 5 feet, and at the same 
time equidistant on the curve ; that which might be allowable 
in a very flat curve. 

The conditions 2 B D nM — o and 2 ^nMx — o show H" K" 

to be the line of action of the horizontal force, which will be 
called T h . The line H" K" is the same line as H" K" in PI. 
V., since it is located by exactly the ‘same condition, the 
vertical intercepts between it and the curve BAD represent¬ 
ing the moments M. Since the only force acting on the rib 
is T h , the bending moment M will be equal to T h multiplied 
by the proper vertical intercept. Thus the moment at b, will 
be T h x afi t ; and that at b 8 , T h x a 8 b 8 . 

The value of T h is determined by either of the conditions 
2 ^nMy=D h EI, or 2 B j nMx=DEI\ by way of variety the lat¬ 
ter will be taken. Assume any point as C for the pole, and 
make BC the pole distance. Make B 1 equal to ^Aa 9 ; 1 — 2 equal 
to a 8 b 8 ; 2 — 3 equal to a ; .... and 9—10 equal to ab, and 
construct the polygon Ca'a". . . . a xi in the usual way. B — 10 
is one half H'b', and a x Ii' is a vertical line through //, while 
Bb' is one quarter of Bb. Draw the horizontal line a x E , then 
the vertical intercepts included between a x E and the deflec¬ 
tion polygon a x a v C, of the kind (ah), represents in an exag¬ 
gerated manner the deflections (vertical) of the points in the 
rib vertically above them. 

Now since the actual moment in an equilibrium polygon is 
equal to the vertical ordinate multiplied by the pole distance, 
the actual vertical deflection at A is CE multiplied by its pole 
distance; however, since the real deflections are proportional 
to the vertical ordinates of the kind (ah), those deflections 
may be at once found by multiplying those ordinates by a 
proper ratio. That ratio is a known quantity, because the 
real deflection at A is known to be three inches. The ordi¬ 
nate CE measures 1.61 inches on the drawing, and represents 
16.1 feet on the actual rib; the ratio desired is therefore 

3 = 3 = JL_ 

12 x 16.1 193.2 64.4' 


To illustrate, the deflection at the 




THERMAL STRESSES IN ARCHED RIB. 253 

section t is 13.9 ■— 64,4 = 0.216 feet. Similarly, the deflection 
at b z is 6.2 -T- 64.4 = 0.096 feet. These quantities will be used 
farther on. 

It will now be necessary to give a little consideration to the 
general equation EnMx = DEI. 

If the ordinates of the kind (ab), measured from H"K", are 
denoted by y\ M, from what has already been said, can be 
written as T h y' . Hence the general equation may be written, 

EnMx = T h Eny'x = DEL 


Or, since n is constant, 



DEI 

n 'Ey ’ x ' 


This last is the equation from which T h is to be found. 

Now since y is positive or negative according as it is meas¬ 
ured on one side or other of the line H"K" , and since n is 
assumed to be uniform and horizontal, the quantity nEy'x is 
the difference between the statical moments of the moment 
areas on the different sides of that line in reference to the sec¬ 
tion considered. Written as an integral expression, it would 
be fy'xdx. 

The area of the surface Ab 6 ma 9 is 1.12 sq. in., or 112 sq. ft. 
full size. The distance of the centre of gravity of the same 
area from AC is very simply found by construction. Take 
any point a 7 for a pole, and a 7 m for the pole distance. Make 
mm t = mm 5 = ia 7 b 7 , equal to a & b 6 , and so on, making, 

however, equal to a half of Aa 9 . Then construct the 

equilibrium polygon e, e 1} e%, ... . e 7 . The sides ee\ and e 6 e 7 
produced will cut each other in the vertical line GH passing 
through the centre of gravity of the area Ama 9 . 

The area BH 'm is to be treated in precisely the same man¬ 
ner, taking a j as pole, and a 1 H'' as pole distance, and making 
Bn equal to a half of BH". The vertical line KL passing 
through the centre of gravity of the area is found as before at 
the intersection of the sides f/[ and f 5 f G prolonged. 

The area BH"m is, of course, equal to the area Ama 9 , con- 



ARCHED RIBS. 


254 

sequently the value of n'E y x for the section A will be the 
product of the common area by the horizontal distance be¬ 
tween their centres of gravity, i. e ., 3.26 inches in the figure, 
but 32.6 ft. full size. 

The cross-section of the rib will be assumed to be of such 
form that El has the value of 2,000,000 foot-tons. Hence for 
the pointH,Z>=3 inches or 0.25 ft., n'Sy'x = 112x32.6 = 
3651.2; 


_ 2000000 x 0.25 

.-. T h = - -v-—,= 137 tons. 

3651.2 


For any other section, as t, the deflection is 13.9 -f- 64.4 = 
0.216 ft. The vertical line passing through the centre of 
gravity of the area included between m and a vertical line 
through t passes through the intersection of the sides e 3 a^ and 
ee lt prolonged, and the distance between KL and it is, as 
shown, 3.04 inches in the drawing. For this section 

n 2 y'x = 112 x 30.4 — 49.82 x 5.8 = 3116; 



2000000 x 0.216 
3116 


139 tons. 


In precisely the same manner, for the point b 3 , 

~ 2000000 x 0.096 

T h =-— = 135 tons. 

1421 


The values should have been the same, except for the 
errors incident to a small scale and the fact that polygons 
were used where curves really belonged ; yet the difference 
between the extreme values is only 2 \ per cent, of the larger, 
which is not very much of an error. 

If, however, the true value of n (5.47 feet, nearly) be taken 
along the centre line of the rib, a decidedly different result will 
be found, since n 2 y'x will have a different value. 

The quantity x, as before, will be measured from a 9 to the 
intersections of the dotted lines drawn through the points (£), 
while y' will represent any vertical ordinate (belonging to any 





THERMAL STRESSES IN ARCHED RIB. 


point (^)) from the line H'K", taken positive downward. 
The point m will be assumed midway between a s and a A . 

The following values are measured from the original draw¬ 
ing : 


x — 1.36 

y' = - 

- 6.2 

1 (xy') = - 

4.22 

* = 5-45 

y = - 

- 6.1 

xy' = — 

33-25 

x = 10.9 

/ = - 

- 5.5 

X y — — 

60.00 

X — 16.2 

r 

y = - 

- 4.6 

xy — — 

74.52 

X = 21.7 

/=“ 

- 3-2 

xy = — 

69.44 

X — 26.8 

/= - 

- 1-35 

xy' = — 

36.18 




— 

277.6l 

x = 32.0 

y = 

0.7 

xy' — 

22.4 

X — 36.8 

y = 

3-2 

xy — 

I 1776 

X — 41.6 

/ = 

6.0 

xy’ — 

249.6 

X — 46.0 

y = 

9.2 

xy' — 

423.2 

x — 49.0 

y = 

11.98 

II 

> 

293-5 


1084.06 


The first values of x and y belong to that portion of the 
moment area adjacent to Aa 9 . 

The last values of x and y belong to that portion of the 
moment area adjacent to BH"; and half of each product is 
taken, so that it may be multiplied by the full value of n — 
5.47 feet. 

The formula then gives : 


_ DEI 
h n^y'x 


0.25 x 2000000 
4411.30 


113.3 tons (nearly).* 


* The method by deflection polygon given in the next Article, produces a re¬ 
sult essentially the same as the one above. 

BC is the pole distance laid down to a scale of 400,000 foot-tons to the inch, 


T h = 


4 X 16.1 x 5.47 


X 


0.62 x 400000 
6.2 


113.6 tons. 


The agreement is much closer than can ordinarily be expected with the scales 
used. 









ARCHED RIBS. 



ft 


The difference between the results of the two methods is 
137.00 — 113.3 = 23.7 tons. 

This difference is by no means small, and shows how care¬ 
fully the approximate method ought to be used. 

With a wrought-iron rib, the deflection taken, 3 inches at 
the middle of the span, belongs to a change of temperature 
of about 165° F., a very extreme case, which accounts for the 
large values of T h . 

This shows, however, in a very marked manner, the impor¬ 
tance of putting together an arched rib at about the mean 
temperature, for then the variation of temperature to be taken 
in the calculation of thermal stresses will only be about half 
the variation between the extreme limits. 

The effect of T h is the same as if that force were applied at 
m and acting toward H" for the portion Bin fixed at B, but 
applied at m and acting toward ni for the free-end portion 
(so considered) mA. 

The change of temperature, 165°, changes the radius from 
75 feet to 74.536 feet, and increases the length of the curve 
0.122 feet. 


Art. 51.—Thermal Stresses in the Arched Rib with Ends Free. 

The method to be used in the present case is somewhat 
shorter and simpler than the one used in the preceding, but 
will probably not give as nearly correct results when the scale 
used is small. 

The figure to be used is that shown in PI. VIII., and the 
curve BAD is precisely the same as that shown in Pis. V., 
VI., and VII. 

For the sake of greater accuracy, the curve BA will be di¬ 
vided into ten equal parts. There will then result, n — bb Y — 
b x b 2 = etc., — 5.47 (nearly) feet. 

Since BD is the true position of the closing line for free 
ends, the effect of the variation in the temperature will be 
the same as that of a horizontal stress, T h , whose line of 
action is BD, and which will produce a deflection equal to 
the thermal deflection. What may be called the “thermal 


THERMAL STRESSES IN ARCHER RIB. 


25 7 


moment,” therefore, at any point will be equal to T h multi¬ 
plied by the vertical ordinate of the curve at that point, or 
M — T h y'. The moment at b 5 , for instance, is M = T h x b 5 n 5 . 

Supposing the rib to be of wrought iron, a change of tem¬ 
perature of 137° • will cause the length to change from 

io 9-454 f ee t to 109.556 feet, and the radius from 75 feet to 
74.67 feet; the corresponding upward vertical deflection at 
crown A will be ij inches, or ^ of a foot. 

Let the two general equations be compared: 

2 Px = T h y. 

2 nMx = l 2 M'x = El. D. 

From these two equations it is seen that if M ' be taken as 
vertical loading, and El as pole distance, then the ordinates 
of the resulting equilibrium polygon will represent the deflec¬ 
tions according to tJie same scale by which x is measured. It is 
important also to notice that M 1 and El are of the same de¬ 
nomination (foot-pounds or foot-tons, as the case may be), 
consequently M' is to be measured in the same scale as that ac¬ 
cording to zvhich El is laid dozvn. 

Since nM — nT h y' ( n being constant), the vertical ordinates 
of the kind (bn) are proportional to the moments M or M 
and they may be taken to represent those moments ; since, 
however, that would carry the lower extremity of the load 
line B — 10 off the diagram, one third only of those ordinates 
will be taken in the plate. 

As before, El will be assumed to be 2,000,000 foot-tons, 
and the scale according to which it is to be laid down for the 
pole distance at 200,000 foot-tons to the inch. Hence make 
EHF parallel and equal to BCD. Since BD is ten inches in 
the diagram, EF is the pole distance, and F will be taken as 
the pole. 

As in the case of the same rib with external loading, the 

71 

half intercept at the distance — from B will be neglected. 

Make E 1 equal to J (A AC ), 1-2 equal to lb 8 n 8 , etc., and 

construct the polygon a a l a 2 . H in the usual man- 

17 



258 


ARCHED RIBS. 


ner. BA is divided into ten equal parts for the sake of greater 
accuracy. 

The polygon thus constructed will represent the actual de T 
flections to a scale of io feet to the inch. 

Now Ea is equal to 1.12 (3.32 inches on the original draw¬ 
ing) inches, or 11.2 feet full size, whereas it ought to be but 
0.125 foot; and since the pole distance is to remain the same, 
the moments must be reduced in the ratio of J to 13.2 ; or the 


moments as actually taken must be multiplied by 


8 X 11.2 


89.6 


. The reduction for the bending moment at A, there- 


, . . . ^ n . 0.636666 x 200000 . . ~ 

fore, is l AC x 200000 -f- 89.6 = — --— p -, since AC 

3 y 89.6 

is equal to 1.91 inches. Hence, 



0.636666 x 200000 
89.6 x 19.1 


= 74.46 tons. 


But since M' — nM : 



M’ 

ny 



13.6 tons. 


This operation may be considerably shortened by remem¬ 
bering that 0.636666 -f- 19.1 = 1 30, and that this ratio is 

constant for all points. If, therefore, the moment at any 
other point, as b$, he taken, precisely the same result will be 
obtained. 

The method used in the preceding Article gives, at least 
approximately, the same result. Taking Bn , Bn x , etc., for 
the different values of ;r, and bn, b x n x , etc., for y, there will 
result : 


x = 4.0 feet, 

y = 

37 f eet > 

00 

• 

HH 

II 

> 

x — 8.4 “ 

y = 

6.8 “ 

xy — 49.6 

= 13.2 “ 

r 

y = 

9.5 u 

xy — 101.6 

x — 18.O “ 

r 

y = 

12.0 “ 

xy — 183.6 









ARCHED RIBS WITH FIXED ENDS. 


x — 23.1 feet, 
x =. 28.2 “ 

X = 33-6 “ 
x — 39.0 “ 

* = 44.4 “ 


y — 14.2 feet, 
y = 16.0 “ 

/ = 174 “ 

y = 18.3 “ 

y = 18.9 “ 


xy = 307.2 
xy =425.8 

xy’ = 554-4 
xy = 643.5 

xy — 799-2 


With these values the partial summation is : 

2 ny'x — n^y'x = 16846.00. 

A product for the moments adjacent to B is nearly : 

0.92 x 1 x 2.73 = 2.51. 

Another for those adjacent to Ad' is nearly : 

19.1 x 48.64 X 2.73 = 2536.5. 


Taking the sum of these results for the complete summa¬ 
tion : 


DEI 0.125 x 2000000 

T h = ^—— = — q - =13.0 tons (nearly). 

n 2 >nyx 19385.00 J v 


The polygon a , a ly a 2 , a 3 , . ... H is an exaggerated repre¬ 
sentation of the movements of the points b } b ly b 2 , etc., when 
the temperature is changed 137 0 F. 

In all cases, as large a diagram as possible must be used, in 
order to reduce the scale for the pole distance, so that that 
distance may be the largest possible. 

The use to be made of T h will be shown farther on. 


Art. 52.—Arched Rib with Fixed Ends—/ and n Variable. 

The rib to be taken in this case, and the loading, are pre¬ 
cisely the same as taken in the preceding cases. As before, 
the centre line is divided into ten equal parts of 10.95 feet 
each. The data are therefore the following: 

Span = 100.00 feet. 

Radius = 75 -°° “ 

Panel fixed load = 4.00 tons. 

Panel moving load = 10.00 “ 

Centre rise of rib = 19.10 feet. 





26 o 


ARCHED RIBS. 


The moving load is supposed to cover the half span BC. 
The figure to be referred to is that shown in PL IX. 


The point h is midway between b' and b ", while k is mid¬ 
way between b" and b'". 

The moment of inertia / of the cross-section of the rib will 
be taken as 2,000,000 foot-tons throughout Bh, 1,777,778 foot- 
tons throughout hk, and 1,600,000 foot-tons throughout kA. 
The same values hold for similar portions of AD. 

1, 2, 3, etc., .... 10, is the load line, and BC the pole distance 
of the equilibrium polygon Ea 2 Ca^F. That polygon is drawn 
in precisely the same manner, in fact, is precisely the same 
one as that shown in PI. V.; the radiating lines drawn from 
C are, therefore, omitted. As the ends are fixed, EF cannot 

B flJVt 

be the closing line; but the condition 2 . -=-= = o must first 

n El 


be imposed. 

It has already been shown that for any point in the arched 
rib the moment M = M a — M b . 

Hence if I x be the moment of inertia for the portion 2 Akoi 
the rib, there may be written : 


v snM 
*>-£/ 



niM a — 2 b b niM^j — o. 


For Bh , i has the value j = 0.8 ; for kh, the value ^ 
and for Ak, the value y 1 = 1.0. 

A 

The closing line UK must be so located that: 


0.9 ; 


2 * niM a - o. 


But for the equilibrium polygon Ea x a^ etc., the summation 
2 b d niM a has the value : 

2 D niM a — o.Sn' x vv + 0.8;/ x a x c + 0.9^2 x a^c r — n x 
c r "a 3 — n x c iy a 4 — n x c x C — n x c yV a & — n x c viI # 7 — 0.9 n x 
c ym a 3 + o . 8 n x + 0.8 n' x v'"v"= o. 



ARCHED RIB WITH FIXED ENDS. 


261 


On the curve BAD the distance Bb is — = J x Bb\ and H'v 

4 

is taken vertically through b ; also Db x — Bb and K'v'" is 
drawn vertically through b x . From the location of b and b x it 

follows that ri — —, consequently 71 may be canceled from the 

w' v '" v " 

series by writing — and -instead of the whole quantities 

2 2 

themselves. 

If vv be taken at 1.04 inches and v'"v" at 0.66 inches, the 
above summation (after dropping n in the manner shown) 
gives a result of — 0.02 of an inch only. This agreement is 
sufficiently close. 

The vertical intercepts and their products by i, are the fol¬ 
lowing : 


+ 0.8 


T 0.8 

+ 0.9 

— 1 

— 1 

— 1 

— 1 

— 1 


+ 0.9 
+ 0.8 

+ 0.8 


vv 

X - 

2 


+ 0.8 


x a x c = + 0.8 
x ci^c = + 0.9 

tn _ 

x a s c = — I 
x a±c u = — 1 
x Cc y = - 1 

X = — 1 

x a 7 c vii = — I 
x a 8 c vlil = +0.9 
x a 9 c lx = + 0.8 

it ut 

V V ' , 0 

x-+ 0.8 

2 


1.04 

X -— = 

2 

x 0.56 = 

x 0.06 = 
x 0.30 = 
x 0.50 = 
x 0.46 = 
x 0.31 = 

X 0.10 = 

X 0.15 = 
x 0.42 = 
0.66 

x-= 

2 


+ 0.42 

+ 0.45 

+ 0.05 

- 0.30 

- 0.50 

- 0.46 

- 0.31 

- 0.10 
+ 0.13 
+ 0.34 

+ 0.26 


B • 

The other condition for the closing line HK is 2 D nM a ix = o. 








262 


ARCHED RIBS. 


Taking the products of the ordinates ( nMi ) of different signs, 
by their horizontal distances from DK" , as was done in Art. 
48, there will result the sums 4- 91.8 and —90.5. The alge¬ 
braic sum is only + 1.3, which is near enough to zero. HK 
will therefore be taken as the proper closing line. 

The condition which locates the closing line H"K” is 
similar to that which placed HK ; it is the following: 

2 ^niM b — 0.8 x 2 n x H'b 4- 0.8 x 211 x c x b' 4- 0.9 x 2 n xc 2 b " 
— 2 n x b f "c 3 — 2 n x b iy c A — n x Ac 5 = o. 

Since the curve BAD is symmetrical in reference to AC, 
the line H"K" will evidently be horizontal. For the same 
reason, 211 ' and 2 n are written in the summation in ail its 
terms except the last. As before, n is one half n, and by so 
writing it, n may be dropped from the series. 

By making Ac 5 = 0.58 inch, the summation (after dropping 
71) gives 2.18 — 2.16 = 0.02 only; H"K" will therefore be 
assumed to be the proper closing line for BAD. 

The vertical intercepts and their products by i are the fol¬ 
lowing : 

o bH' 0 1.13 

2 x 0.8 x —- = 2 x 0.8 x —- = 0.90 
2 2 

2 x 0.8 x b'c x = 2 x 0.8 x 0.66 = 1.06 

2 x 0.9 x b"c 2 — 2 x 0.9 x 0.12- 0.22 

2 x b ' c^ = 2 x 0.27 = 0.54 

2 x b iv c± = 2 x 0.52 = 1.04 

Ac 5 = 0.58 = 0.58 

As before, since the curve BAD is symmetrical in reference 

to A , the two conditions 2 B D niM h — o, and 2 B D niM b x = o, are 
equivalent. 

The equation expressing the condition that the horizontal 
deflection of D in reference to B is nothing, is the general 
one already given: 




ARCHED RIB WITH FIXED ENDS. 


263 



nMy 

~FT 


n \ 

Eh 


2 B D riMy = 


n x 

El\ 



Or, as in preceding cases : 

2 S D riM a y = S B D riM b y. 

The quantity n t is any standard value of n, just as I x is a 
standard value of /, and r is such a variable ratio that for any 

n 

section n — rn x or r = —, In the present case ru will have 

Hi 

the value 10.95 feet; consequently r will be unity for all 
sections except b, and for that one it will be J. 

The ratio r might have been used in the previous summa¬ 
tions of the present Article in exactly the same manner and 
with exactly the same values as in the present one. 

The ratio i has the same value as before. 

The principles on which the remaining constructions are 
based are precisely the same as those shown in the preceding 
Articles; the difference in the construction itself is simply 
this, that riM a and riM b are taken instead of M a and M h . In 
other words, r and z, in general, in this case, have values dif¬ 
ferent from unity, while in the preceding cases r — i — 1. 

The following are the values of ( riM a ): 

x 0.8 .( vv 4 - v’v"') — At; 

I x 0.8 . ( a x c + tf 9 d x ) = //'; 

I x 0.9 . ( a 2 c" — a 8 c yi:i ) = t't" ; 

ix I . (— a % c n — a 7 r vli ) = — t"t " f ; 

I x I .(-V v “ ^ vl ) = - t’"F; 

1 x 1 x Cc y — t iy A. 


Draw radial lines from C to the points t ; then draw the 
horizontal lines bC", b'd^d ’, V'd^d", etc. C"d( is parallel to 
Ct ; di'di' is parallel to Ct '; d^dF is parallel to Ct", etc. 
CC"d 1 , d 1 , 'd 1 n, d 1 lY d 1 y is then the deflection polygon for the mo¬ 
ments M a . 





264 ARCHED RIBS. 

The following are the values for ( riM b ) : 

| x 0.8 x 2 H'b — As; 

1 x 0.8 x 2 b'c x — ss ; 

1 x 0.9 x 2 b”c 2 = s’s" ; 

ix 1 x (— 2 b" f c s )= —s"s f "; 
ix ix (—2 b {y c^)— — s"'s iy ; 
ix ix (— Ac 5 ) = —s iv A. 

The point A belongs to the curve BAD. 

The sides of the deflection polygon C"d'd ,, d" , d iy d y are 
parallel to radiating lines drawn from C to the points s. 

Ad\ represents 2 D riM a y , and Ad y represents 2 D riM b y. 

Since the first is less than the second, the moments M a must 
be increased in the ratio of Ad y to Ad y . Hence on AC pro¬ 
longed, make CM equal to Ad y ; draw MN parallel to BD 
and with a radius CN equal to Ad y find the point N. Pro¬ 
long the line CN ; this line will enable the true moments M a 
to be determined in the manner already shown in the other 
cases. 

Draw c y p parallel to BD, c 5 a y equal to Cp will give a point 
a y in the true equilibrium polygon. Again, Ck" equals KF, 
hence K"a x , equal to Ck"\ gives the true point a x . 

Also, Cg is equal to HE ; and H"a, equal to Cp’, gives the 
true point a. All points in the true polygon might be thus 
determined, but it is advisable to check by the other method 
already shown. 

For this purpose draw CL parallel to HK, and LC' parallel 
to BD. Then make CP equal to BC and draw OP parallel to 
BD. Take LC 1 equal to CO. C' is the pole and LC' the 
pole distance of the true equilibrium polygon. Finally draw 
radiating lines from C' to the load points 1, 2, 3, 4, 5, 6, 
etc. Starting from any point already determined, as a , draw 
aa! parallel to C' 10; da" parallel to C'g; a'a" parallel to 
C'8, etc. The different points found by the two methods 
ought to coincide. 

The polygon ada'd"a iy a y a yi a yil a ym a ix a x is the true equili¬ 
brium polygon, which was to be found. 


ARCHED RIB WITH FIXED ENDS. 



C'L is 3.86 inches ; hence T h , whose line of action is H"K" 
is equal to 3.86 x 15 = 57.90 tons. 

In the determination of the thermal stresses the same fig¬ 
ures will be used, and there will be supposed such a change 
of temperature that the point A will suffer a vertical deflec¬ 
tion of 1.5 inches or 0.125 of a foot; the same, in fact, as was 
supposed in a previous case. 

As the ends of the rib are fixed, the general conditions, 
2 ^riM = o; and 2 ^riMy = 2 ^riMx = o hold as well for 

thermal stresses as others. Consequently H 'K” will be the 
line of action of the horizontal stress T h , induced by the 
variation of temperature. 

As before, let y denote the vertical ordinate of any point 
in the curve BAD from the line H 'K'', then there may be 
written : 

#1 -^A . n/r 7 ^ #1 L h A • , 

2 ^nMx — D — nT - ^ riy x 

EL D EI X D 


T h = 


DEL 


.A • t 

7 i x 2 . D ny x 


In order to save confusion in the figure, n x will be taken at 
its previous value 10.95 feet. 

Also, 

EL — 1600000 foot-tons ; 

D = 0.125 foot. 


A half of the moment at A will be supposed applied at a 
point e distant —- from A on the curve BAD , and none at 
all at A. 

The co-ordinate will be measured from c 5 towards K'\ 
The following values then result: 

* = 2.72 ft. riy' = \ x 1 x 5.7 = 2.85 riy'x = 7.752 

* = 10.90 “ riy' = 1 x 1 x 5.2 = 5.2 riy'x = 56.68 

x = 21.7 “ riy' - 1 X 1 x 2.7 = 2.7 riy'x = 58.59 


123.022 






266 


ARCHED RIBS. 

\ 


- 

x = 3.2 ft. 

riy' = 

— I x 0.9 X 

1.2 = 

— 1.08 

riy x — 

- 34-56 

* = 41-6 “ 

riy’— 

— I xo.8 x 

6.6 = 

-5.28 

riy' x — 

— 219.648 

*=47.9 “ 

riy'— 

X 

00 

• 

0 

X 

1 

n .3 = 

-4.52 

riy x — 

— 206.508 







—460.716 


Hence, 

2 A D riy'x — 123.022 — 460.716 = — 337.694. 


The negative sign will be dropped hereafter, as it refers 
simply to the direction in which y was measured. 

Making the substitutions: 


0.125 x 1600000 

10.95 x 337-694 


= 54.1 tons. 


The method by the deflection polygon gives nearly the 
same result, as will now be shown. 

Comparing the two equations: 


-2 Px = T h .y, 
'EnyiMx — EI X . D, 


it is seen that if EI X be taken as the pole distance in the 
deflection polygon, n x riM must be the general expression for 
the load at any point. 

Since EI X is 1,600,000 foot-tons, CD will represent it at 
320,000 foot-tons per inch; and that will be taken as the pole 
distance. D — 3, measured downwards, will be the load line. 

If the loads were taken at 10.95 x riM, or 10.95 riy', the 
lower limit, 3, of the load line would not be on the diagram. 
The loads will therefore be taken as A^riy and a proper reduc¬ 
tion will be made afterwards. Hence, make 


D 

- 1 = 

4 riy' = 

4 x 0.285 inches. 

1 

— 2 — 

4 riy' = 

4 x 0.52 “ 

2 

- 3 = 

4 riy' — 

4 x 0.27 “ 

3 

-4 = 

— 4 riy' = 

— 4 x 0.108 “ 

4 

- 5 = 

— 4 riy' = 

— 4 x 0.528 “ 

5 

-D = 

— 4 riy' = 

— 4 x 0.452 “ 




STRESSES IN THE MEMBERS. 


267 

The values of riy' are taken from the table immediately 
above. 

By drawing radial lines from C to the points 1, 2, 3, 4, 5, 
the polygon Cd^d^d^d^d^d^d^ is formed in the usual manner. 
The deflection Dd\ measures 2.67 inches or 26.7 feet, full 

size. Since — = 2.74, the deflection with the true moment- 
4 

loads would be 26.7 x 2.74 = 73.16 feet, whereas it should be 
but one-eighth ot a foot. Hence, measured by the same scale, 

the quantities n x nM must be —-?-- of those taken in the 

8 x 73.16 

figure, the pole distance CD remaining the same. 

Since M — T h y, there may be written for the points : 

10.95 x riT h y' = 10.95 x 0.285 x 320000-f- 8 x 73.16. 

As r — i, i — 1 and y' = 5.7 feet there results: 

Tfi = 54.7 tons. 

The deflection at other points might be used in the man¬ 
ner already shown in a preceding Article. 

It is thus seen that the constructions are equally simple in 
principle whether r and i are constant or variable. 

If the ends had not been fixed, it would only have been 
necessary to use the condition 

n \ ^ b . n/r 

etC u riMy = 


Art. 53.—Determination of Stresses in the Members of an Arched Rib— 

.» 

Example—Fixed Ends—Consideration of Details. 

It has been shown, in the preceding Articles, how to deter¬ 
mine the horizontal tension T h in the various cases which 
may arise; the method of using this horizontal tension in the 
determination of the stresses in the individual members of an 
arched rib remains to be shown. 

For this purpose there will be taken the rib shown in PI. 




268 


ARCHED RIBS. 


X., Fig. 3, having ends free, i. e., free to turn about the points 
M and L. 

The curve which has hitherto been used, and called the 
“ centre line” of the rib, is the centre line of the neutral sur¬ 
face of the arched rib considered as a beam ; consequently the 
centres of gravity of the various cross sections of the rib must 
be found in this “ centre line,” in all cases. In other words, the 
“ centre line ” which has been used in the preceding articles 
is the locus of the centres of gravity of the normal cross sec¬ 
tions of the actual rib. 

In the rib taken as the example, PI. X., Fig. 3, the apices 
in the upper and lower chords lie in the concentric circum¬ 
ferences of circles having radii of 78 and 72 feet, respectively ; 
and the centres of gravity of the normal cross sections will be 
supposed to lie on the circumference of a circle having the 
same centre, whose radius is 75 feet. Thus the centre line is 
precisely the same as has been used in the preceding Articles. 
The same span and loading will also be taken. 

The extremities of the span, or points M and Z, at which 
the horizontal tension or force is applied, must lie in the 
centre line. 

All the loading will be assumed to be applied at the apices 
of the upper chord, although the operations would be exactly 
the same if the fixed load were divided in any proportion be¬ 
tween the two chords. 

The apices of the triangles of the web system were located 
as follows: As was done in finding T h , the centre line was 
divided into ten equal parts. The upper chord panel points 
are vertically over those points of division. The upper chord 
panels were then bisected, and radii were drawn through these 
points of bisection. The lower chord panel points were taken 
at the intersections of those radii with the circumference of 
the circle whose radius was 72 feet. 

There is only one point to be observed in forming the chord 
panels, i. e., the panel points must be so located that the load 
will act exactly as was supposed in determining T h . 

The following loads will then be assumed to act through 
the upper chord panel points: 


STRESSES IN THE MEMBERS. 



7 

14 

9 

4 

2 


tons at the intersection of 1 and 3 ; 

“ “ intersections of 4 and 5, 6 and 7, 8 and 9, 10 

and 11 ; 

“ “ “ intersection of 12 and 13; 

“ “ intersections of 14 and 15, 16 and 17, 18 and 

19, 20 and 21 ; 

“ “ “ intersection of 22 and 24. 


As the ends are free, L — 10 = 50.7 tons in PL VI., gives 
the reaction R at the left end of the span, or at L in the ex¬ 
ample. Also L — 1 in PI. VI., gives the reaction as 30.3 tons 
at M in the example. In Art. 49 it is found that T h — 52.5 
tons for this case, and the two methods in Art. 51 give the 
thermal stresses in the horizontal tie as 13.6 and 16.7 tons. 
The thermal tension will be taken at 15 tons. The total ten¬ 
sion in the tie will then be 52.5 4 15.00 = 67.5 tons, as 
shown. 

It is a matter of no consequence whether the tie exist or 
not. If it does not exist, the abutments at M and L must 
then supply the horizontal force of 67.5 tons. 

Fig. 4 of PI. X. is the complete diagram for the stresses 
with the load taken ; it is drawn to a scale of 20 tons to the 
inch, nearly. The lines indicated by letters or figures in the 
diagram are parallel to the members of the rib indicated by 
the same letters or figures, though the parallelism is not ex¬ 
actly shown in the plate for all the lines. 

With the following explanations relating to the diagram, 
little more is needed : 


a'b' — b’c' — cd' — d'f — 14 tons. 
011 — 11771 — trik — k'h =4 “ 

*/' = 9 “ 

q'd = ( l) + 7. 

P o' = (24) + 2. 

PS = {E ). 


ae is the reaction R , and e’o' is the reaction at M f while 


2 VO 


ARCHED RIBS. 


e P is the horizontal tension 67.5 tons. Although the diagram 
appears very complicated, yet, it is really composed of very 
simple yzW-sided figures, as may easily be seen. Let the rib 
be divided through c, 7, B, and T ; then the portion of the rib 
between that surface of division and L is held in equilibrium by 
the action of the stresses in the members divided (considered 
as forces external to that portion), the applied loads, ( T ), and 
the reaction R. The resultant of the loads and R is a verti¬ 
cal shear represented by ec in the diagram. The forces act¬ 
ing upon the portion of the rib in question are then repre¬ 
sented by the lines ec, (T), ( B ), (7), and ( c ) in Fig.4, and these 
constitute a simple five-sided figure. The arrow heads show 
the direction of action of these forces, and enable the kind of 
stress to be recognized at a glance. 

The whole diagram is thus composed of just such pen¬ 
tagons. 

As a check on the accuracy of the construction of the 
diagram, if it is worked continuously from L to M, the four¬ 
sided figure involving (T), (23) and (24) should exactly close. 
It is far more conducive to accuracy, however, to work up 
the diagram from both ends, and if the work has been accu¬ 
rately done, the diagrams will give the same stress in that 
member which becomes common to both where they meet. 

The stresses, as determined by the original diagram from 
which Fig. 4 was constructed, are written in Fig. 3. 

If an arched rib is subjected to a load, advancing panel by 
panel, the stresses due to the fixed load alone may first be de¬ 
termined and then tabulated. The stresses in all the mem¬ 
bers of the rib due to each panel moving load may then be 
found and tabulated also. The greatest stress of either kind 
in any member may then be determined by a combination of 
these results in the usual manner. 

Some of the stresses found by diagram should be checked 
by moments in the following manner. The horizontal dis¬ 
tances of the panel points, in the left half of the rib holding 
14 tons each, from a vertical line bisecting the span and pass¬ 
ing through the intersection of 12 and 13, are 10.9 feet, 21.7 
feet, 32 feet, and 41.5 feet. The normal distance from 


S7RESSES IN THE MEMBERS. 


271 


the intersection of (12) and (13) to E, is 6.15 feet (by scale). 
Hence, taking moments about that intersection, 

/ _ 50-7 x 50 - 14 (10.9 + 21.7+ 32 +41-5) ~ 67.5 x 22.1 

1 j 6j 5 “ “ 

— 72 tons. 


The diagram gave 72.5 tons, and the agreement is suffi¬ 
ciently close. 

On account of the ill-defined intersections of the prolonged 
chord sections in any panel, the method of moments for the 
web stresses is not satisfactory unless one chord stress in the 
panel is known. The web stress can then be found by mo¬ 
ments in a manner to be presently illustrated. 

Again, let ( G ) be determined by taking moments about the 
intersection of 16 and 17. Draw a vertical line through that 
point. The horizontal distances of the three upper chord 
panel points on the right of that line, from the same, are 
10.3 feet, 19.8 feet, and 28.5 feet (by scale). In the same 
manner the vertical distance of the point above T is 19.25 
feet. Hence, 


( r\- 3°-3 x 28.5-4(10.3+ 19.8)-67.5 x 19.25 _ 

6.15 


— 90.5 tons. 


The diagram gave 95 tons, and the agreement is not close. 
This illustrates in a marked manner the great fault of the 
graphical method. In constructing the original diagram, 
shown by Fig. 4 of PI. X., the rib was drawn to a scale of 5 
feet per inch, and the diagram itself to a scale of 10 tons per 
inch, and although the greatest care was taken, yet the stresses 
found for the right half of the rib may, in some members, be 
wrong to the extent of even twenty per cent. The method 
requires the largest and most accurate figures possible, and 
the very nicest instruments, for extended diagrams. 

By far the most accurate, and, all things considered, the 
most satisfactory method, is the combination of moments and 






272 


ARCHED RIBS. 


diagram, so freely used in the treatment of bowstring trusses. 
In this method a stress in either chord is found by moments, 
the other two stresses (one a chord and the other a web) in 
the same panel are then immediately found by a simple five¬ 
sided figure or diagram. 

The chord stresses ( E ) and ( G ) have just been found. In 
Fig. 1 take bd equal to 67.5 tons and parallel to T, and make 
dc (parallel to E as well as T) equal to (E) = — 72.00 tons. 
A section is supposed to be taken through the panel in which 
f, 13 and E are found; consequently the vertical shear S = 
50.7 — 65.00 = — 14.3 tons. The remainder of the diagram 
needs no explanation. It gives: 



(13) = — 20.3 tons ; (/) = + 17.4 tons. 

Fig. 2 is drawn in precisely the same manner by using 
( G ) = — 90.5 tons, which has already been determined by 



the moment method. The section is taken through G , 17 
and h. 

Fig. 2 gives: 

(17) = — 5.5 tons ; (//) = + 23.0 tons. 

All the other stresses may be found in the same manner. 
The difference in the case of (Ji) between the results of the 





STRESSES IN THE MEMBERS . 


273 

two methods is four tons, or about 17 per cent, of the smallest 
result. 

By the method of Figs. 1 and 2 accurate results may be 
obtained by taking a scale of even twenty tons to the inch, 
but a larger diagram is preferable. 

If the ends of the rib are fixed, the shortest method of find¬ 
ing the stresses is in no way different from that given in con¬ 
nection with Figs. I and 2, excepting this: the chord stress 
which is found by moments will have a different value. If 
the ends are fixed, the reaction R, at tJie left end of the span, 
will be L — 10 of PI. V. ; and the horizontal tension ( T ) will be 
taken as 54.8 4 - 113.3 = 168.1 tons, from Arts. 48 aiid 50. 

The line of action of T h for both external load and thermal 
stresses is Ed" K " of either PI. V. or PL VII.;, let PI. V. be 
considered. 

The action of T h through H" (taken as acting toward K "), 
so far as the rib BAD is concerned, is equivalent to T h acting 
through B toward D, combined with a right hand couple whose 
force is T h and whose lever arm is BH " . Let the moment ot 
this couple be called M. This moment will cause compression 
throughout the upper chord of the rib and tension throughout 
the lower. 

Let M x represent the moment (of the external forces and 
T h ) about any panel point of the rib, as F, Fig. 3 (the reaction 
and T h being taken for the particular case, as just indicated). 
Then any chord stress, as (BP), will be 

M+ M x 

__ • 

f 

7/1 



n x being the normal depth of the rib, as shown. Particular 
care is to be taken in regard to the signs of M and M x , i. e., it 
is to be noticed whether they tend to produce the same or dif¬ 
ferent kinds of stress in BE. 

After (BE) is found, the diagrams are to be drawn precisely 
like Figs. I and 2, and the resulting stresses, scaled from the 
diagrams, are the ones desired. 

The following, but longer methods, may also be used; 

18 



ARCHED RIBS. 


274 


The first portion of the operation is simply the application 
of the method by diagram, or the combination of moments 
and diagram, already given in connection with the case of 
ends free. 

All the individual stresses in the rib are to be found in this 
manner; those for the zveb members are the true web stresses 
desired if the rib is of uniform normal depth. The chord 
stresses thus found are, however, in all cases, to be modified. 

Let the normal depth of the rib at any section be n x ; the 
distance BH", PI. V., h" ; and the general expression for any 
chord stress due to the moment M — T h h", c. 

Then will result: 

M T h k" 

n 1 


Then let (<:) be the general expression for any chord stress 
already found without considering the moment M ; the nu¬ 
merical value may be either positive or negative. Finally, 
the resultant chord stress desired will be, for the upper chord, 


and for the lower, 


T h" 

(c) - c=(c)~ 


(c) + c — (c) 4- 


n. 


If the rib is of uniform normal depth, i.e., if n x is constant, 
the web stresses will not be affected by the moment M, for it 
(the moment M ) will cause uniform chord stresses throughout 
the rib. 

If the normal depth, however, is not constant, the moment 



M will cause web stresses which may be determined very ac¬ 
curately in the following manner: 









STRESSES IN THE MEMBERS. 275 

Let it be desired to determine the stress in the web mem¬ 
ber DF of a portion of an arched rib, shown in Fig 3, and let 
w denote that stress. The stress in DE is c, found by the 
method just given. 

In Fig. 4, take (DE ) = c and parallel to DE in Fig. 3. The 
lines (GF) and ( DF ) in Fig. 4, are then parallel to (LFand 


(ni:) - <? 


-yCDF) r W 


(GF) 

Fig. 4. 

DF in Fig. 3, and they are the stresses in those members. 
All the web stresses and the chord stresses in one chord may 
be thus found. This operation is simply the method of Figs. 
1 and 2 applied to this case. 

The web stresses may be found by using moments, only, in 
the following manner : 

Take A as any convenient point in GF. Let l x represent 
AB, and / 2 , AC; these lines are normal to DE and DF re¬ 
spectively. 

Let M be still considered right-handed and positive, and 
let it first be assumed that c is compression. Moments about 
A give : 

—cl x + T h Ji 


w — — 


4 


If c is tension, or belongs to a panel in the lower chord, 
then ( w ) will be the stress in a member like DG. There will 
then result : 


w — 


- Cl \ + T h h" 

4 


In either of these formulae, (w) will represent tension or 
compression according as the result is positive or negative. 

Let ( w ) represent any web stress already found by neglect¬ 
ing M, then will any resultant web stress desired be: 

t 

(w) + w; 

the signs of both these quantities being implicit. 


\ 







276 


ARCHED RIBS. 


As the lever arms 4 , 4 , and are scaled from the draw¬ 
ing, the rib should be laid down as accurately and to as large 
scale as possible. 

In important cases these different methods should be used 
as checks. 

A very common system of bracing for arched ribs, although 
a very unsatisfactory one, is that shown in Fig. 5. The web 



members, bB, cC , dD, etc., are normal to the centre line of 
the rib, and are designed for tensio?i only. The other web 
members are for compression only 

Let the ends be supposed free, and take A OP for the true 
equilibrium polygon for a given load. 

For any given loading the stresses in the different members 
are indeterminate, unless about half of the compression web 
members are neglected. 

With the assumed position of the equilibrium polygon A OP, 
for instance, it is seen that compression will increase in the 
upper chord from b to d (nearly); from that point to g 
(nearly) it will decrease. The compression in the lower chord 
will increase from G to L (nearly) and then decrease from 
that point to N. The points of greatest chord stresses of 
either kind are those at which the polygon and centre line of 
rib are parallel. 

From these considerations it results that the web members 
bC, cD, De , Ef , Eg, Gk, Hk , Kl, IM, and mN may be omitted ; 
they must be omitted, in fact, if the stresses are to be deter¬ 
minate. Having made these omissions, the stresses are to 
be found by the methods already given, as those methods 
are perfectly general . 







STRESSES IN THE MEMBERS. 


2/7 

Precisely the same observations apply to the case of fixed 
ends, or to that in which the normal members are in com¬ 
pression and the others in tension. 

It is by no means certain that the stresses thus found will 
really exist in the rib, but the assumption is the best that 
can be made. This system of bracing is, at best, very un¬ 
satisfactory. 

The free ends of an arched rib are sometimes arranged, in 
regard to support, as shown in Fig. 5, PI. X. There are two 
ties or sets of ties, T’ and T'\ instead of one, T h . A and B 
are the points at which these ties take hold of the rib. E is 
the intersection of AB and the centre line, EF, of the rib. 
The span to be used in finding T h for either exter?ial loads or 
thermal variation is the horizontal distance between E and the 
corresponding point at the other end of the rib. T\ T h , and 
T" are parallel to each other, and ac is normal to the three. 

Now if T’ and T" are determinate, there may be written : 


T ' =^.T h ; 


T" = — T h . 
ac 


In such a case the systems of triangulation in the rib may 
be separated, T' will belong to one and T" to the other. 
The stresses may then be found in each system separately, 
and the results combined for the resultant stresses of the rib. 
If the web members may be counterbraced, the resultant 
stresses are thus determinate. It is not certain, however, 
that the tensions T' and T" will have the values given above. 

For the determination of the stresses in the rib, however, 
it is not necessary to resolve T h into T' and T", except for 
the panel ABDC. 

In the case of a design, if the dimensions required by the 
calculations of this Article give a value to the moment of 
inertia I very different from that assumed in the determina¬ 
tion of T h , for either thermal variations or external load, it 
will be necessary to make an entirely new set of calculations 


278 


ARCHED RIBS. 


with another value of /. This must be done until the agree¬ 
ment between the assumed and required values of / is suf¬ 
ficiently close. 


Art. 54.—Arched Rib Free at Ends and Jointed at the Crown. 

Suppose the rib to be represented in the figure. 

Since there is a joint at A, the bending moments must be 
zero at that point; consequently the equilibrium polygon for 
any load must pass through that point. This fact furnishes a 


A 



very simple method of determining T h . Denote by ^Px the 
moment of all the external forces about the joint A ; then, 
since M must be equal to zero, 

S’ Pr 

ZPx- T h (AC) = o. T h =^T- 

In this rib variations of temperature produce no variations 
of stress in BD , except that due to the slight change of A C , 
as shown by the formula above. This, however, is a very 
small quantity, and would ordinarily be neglected. If neces¬ 
sary, it would be allowed for by taking the value of AC at the 
lowest temperature to which the rib would be subjected. 

Other arched ribs are seldom constructed, but they are to 
be treated by the same general methods, precisely, as those 
used in the preceding cases. 






CHAPTER IX. 


SUSPENSION BRIDGES. 


Art. 55.—Curve of Cable for Uniform Load per Unit of Span—Suspen¬ 
sion Rods Vertical—Heights of Towers, Equal or Unequal—Gen¬ 
eralization. 



Fig. i. 

In the figure, let EH'C represent the cable of a suspension 
bridge carrying a load extending over the whole span. In 
the ordinary experience of an engineer, the load carried by a 
suspension bridge cable is nearly uniform in intensity in 
reference to a horizontal line ; so nearly uniform per foot of 
span, in fact, that it is assumed to be exactly so, and such an 
assumption will be made in the present instance. 

The use of the stiffening truss, to be presently noticed, 
makes this assumption essentially true. 

Let ( ED + BC) — l — span ; BH' — h x ; DH‘ — h 2 ; 
w — load for horizontal foot, and let x be measured hori- 

279 











280 


SUSPENSION BRIDGES. 


zontally from H', the lowest point of the cable. The height 
of the highest tower is, of course, h lf and that of the other h 2 . 

The ordinate of any point P is x, the load on H'M is, con¬ 
sequently, W — wx. Draw PK tangent to the curve at P, 
then by the first principles of statics, it is known that the 
direction of the cable tensions at P and FT and the direction 
of W must intersect in one point N. Since, however, w is 
uniform along Xy the resultant direction of W passes through 
N f half way between FT and M. Hence FH‘ — H’K\ or, 
since FK is the subtangent, the abscissa, FH', of the curve is 
equal to half the subtangenty consequently the curve is the or¬ 
dinary parabola. 

Again, it is known that the horizontal component of the 
tension of a cable will be a constant quantity if the loading 
(as in the present case) be wholly vertical; let that compo¬ 
nent be denoted by H. 

Let GNP be taken for the triangle (right angled) of forces 
at P, in which NP represents the cable tension at P, GN the 
load, W — wx and GP the constant horizontal component H. 

Then let AP be drawn normal to the curve at P\ the 
triangles AFP and GNPwWX be similar. There can now be 
at once written the relation : 


but 


A F _ FP _ x _ i 
GP GN wx w ’ 


GP = H AF — 


H 


w 


— constant 



Now AF is the subnormal of the curve of the cable, and 
since it is constant, the curve is the ordinary parabola. 

The preceding results may be generalized in a very simple 
and easy manner. 

If any two points, as P and Q , be considered fixed, and if 
the portion PQ of the cable carry the same intensity of load 
iv as before, there will at once result the general case of a 
flexible cable carrying a load whose intensity, along a straight 
line, and direction are uniform. There may then be stated 





PARAMETER OF CURVE. 


28 l 


the general principle : If a perfectly flexible cable carry a load 
uniform in direction and intensity in reference to a straight 
line , the cable will assume the form of an ordinary parabola 
whose axis will be parallel to the direction of the loading. 

This principle finds its application in the case of a suspen¬ 
sion bridge with inclined, but parallel, suspension rods. 


Art. 56.—Parameter of Curve—Distance of Lowest Point of Cable from 
either Extremity of Span—Inclination of Cable at any Point. 

Attending to the figure and notation of the previous Ar¬ 
ticle, the equation of the curve, the origin of co-ordinates 
being taken at H\ is: 

x 2 = 2 py ; 

in which 2/ is the parameter. 

Let BC — Xi and ED — x 2 , then there may be written x 2 = 

2 phi, x 2 = 2 ph 2 and 2x x x 2 — 4pVh x h 2 . 

Hence 

(x 1 + x 2 ) 2 =l 2 =2p (V/i 1 + VA, ) 2 = 2p (h x + 2yVq h 2 +h 2 ) . . . (1). 

p — -—- 7 — --. . (2). 

2(Vh 1 + V h 2 f 2(/q + 2 Vlh h 2 + h 2 ) 


If the towers are of the same height, then h x — h 2 — h and 




Now x t is the horizontal distance from the lowest point of 
the cable to that end of the span at which h x is found, i. e. y 
BC in the figure, while x 2 is the other segment of the span, 
and by the equations immediately preceding: 

= —= - -== 

V hi + V h % 









282 


SUSPENSION BRIDGES. 


*2 


IV h 2 

- - -- — • • 

v/ii + v 4 


If h x = lh ; *1 = *2 = - 


• • (5). 

• • (6). 


Referring to the figure, since KH' — H’F=^y, if t is the 
inclination to a horizontal line, of the curve at any point P, 
then x tan i = 2 y ; 


hence, tan 




At the summits of the towers: 


. • 2h x , . 2 h 2 

tan Zi = —- and tan i 2 = -— 


If h x — h 2 , tan i x — tan i 2 = ^ 


. ( 8 ). 
• (9)- 


Art. 57.—Resultant Tension at any Point of the Cable. 

In the first Article of this Chapter there was recognized the 
general principle that if the loading on a cable is uniform in 
direction, the component of cable tension normal to that 
direction will be constant at all points of the cable. In the 
present case the resultant tension at the lowest point of the 
cable will be this constant component H. 

From Eq. i, Art. 55, H — wAF. But AF is the subnormal 
of the curve, and, from Analytical Geometry, it is known to 
be equal to one half of the parameter, or equal to p (using the 
same notation as before). 

Hence, after taking the value of p from the previous 
Article: 


H — wp — 


wl 2 


wl 2 


2 (V ht + Vh 2 ) 2 2 (/q + 2 v 7 q h 2 + h 2 ) 










LENGTH OF CURVE. 


283 

Let R denote the resultant tension at any point, then by the 
triangle of forces GNP, in the figure: 


R — H sec i 


: = H \/ 


1 + 


4/ 


(2) 

• • • • 


Eq. (2) gives the tension at any point. At the summits of 
the towers there are found : 


Rt = Hi /i + ~j- . 


r 2 = ha/ 


If h x — /i 2 , consequently x i — x 2 ~ -, then : 

2 


( 3 )- 


1 + ***-' (A \ 

I+ .( 4 )* 


l 


TJ _wl 2 TTA / / V 

H — > R^ — R 2 — 1 "1 * • * ( 5 )* 


Art. 58.—Length of Curve between Vertex and any Point whose Co¬ 
ordinates are x and y , or at which the Inclination to a Horizontal 
Line is i. 

The usual expression for the length of a part of one branch 
of a parabola, beginning at the vertex, as determined by the 
integral calculus, may easily be put in the following form, 
denoting by c the length in question : 


c — 


x 2 j 2 y 
4 y \ x 


]/ l+A f + h yp- lo s-(x + 1/ [ + 1 (I) - 


Or, using the values for tan i, sec i, and /, determined in 
the preceding Articles: 

c — — \ tan i sec i + hyp. log. (tan i + sec i)\ . . . (2). 














284 


SUSPENSION BRIDGES. 


The total length of the cable will, of course, be found by 
putting x x and h x for x and y in Eq. (i), or i x for i in Eq. (2); 
then „r 2 and // 2 for x and y, or z 2 for i, and adding the results. 

Denoting those results by c x and c 2 the total length will 
then be: 

c x + c 2 . 

An approximate formula sometimes used is determined as 
follows. In the figure of the first Article of this chapter con¬ 
sider H'P to be the arc of a circle, and let .r and y be taken 
as heretofore ; also let R be the radius of the circle. The 
ordinary expression for the length of a circular arc, in the 
integral calculus, is : 



dx 


2 R 2 


X* 


(nearly), 


if is small compared with R. Again, performing the divi¬ 
sion indicated and omitting all terms in the quotient after 
the second, there will result: 



If y 2 be omitted in the expression x 2 = 2Ry — y 2 , and the 
resulting value of R be inserted in Eq. (3), there will at 
once be found : 


x T ^.(4)* 

As before, to find the total length of the curve, x x and k ly 
and x 2 and must be inserted in succession in Eq. (4), and 
the results added. 

If the heights of the towers are equal to each other and to 
h, the total length will be 












DEFLECTION OF CABLE. 


285 

The expressions (4) and (5) are evidently not close approxi¬ 
mations except for very flat curves, in which case the nature 
of the curve is a matter of indifference. 


Art. 59.—Deflection of Cable for Change in Length, the Span Remaining 

the Same. 

The approximate formula (4) of the preceding Article is 
usually used in determining the deflection. 

The total length of the cable is: 

2 (h? h}\ 

Cl + c 2 = x 1 +x 2 +-(— + —), 

3 -*2/ 

Differentiating: 

<tfr + 4> = l(T + r) d ** • • • ( 0 - 

3 


dh = 


3 d (Ci + c 2 ) 




The variation in the length of the cable, whether arising 
from variation in temperature or any other cause, is to be 
put for d (<q -f c 2 ) in Eqs. (1) and (2), then dh will be the cor¬ 
responding deflection of the lowest point of the cable. 

If the towers are of the same height, and, consequently, 


2dci 


16 

T 




dh = 


3/. 

1 6hi 


2 dc x 



* Since h x — h.% — constant, dh x — dh z — dh. 








286 


SUSPENSION BRIDGES. 


It is assumed in Eqs. (i) and (2) that the lowest point of 
the cable remains at the same horizontal distance from the 
ends of the span, though such is not really the case. 

The true deflection can only be found by trial by the use 
of Eq. (1) of the previous Article. 

Let (c x -f c 2 ) be the known length of the cable before varia¬ 
tion in its length takes place ; then let h lf h 2 , x x and x 2 be the 
original heights of towers and segments of span, also known. 
Let y x and_y 2 be the heights of towers above the lowest point 
of the cable after the variation in its length has taken place; 
and let it be assumed, as before, that x l and x 2 remain the 
same whatever the deflection. 

Let v be the variation in length of the cable. 

Then, since v — — (c x + c 2 ) + (<q + c 2 + v) : 



But there is also the equation of condition : 

y\ — — K — K — constant . . . (6). 

The value of y t or y 2 may be taken from Eq. (6) and put in 
Eq. ( 5 )> there will then be but one unknown quantity in the 
right member of that equation, and its value must be found 
by trial. The first value of y x or y 2 taken may be h x or h 2 
increased or decreased, as the case may be, by dh given by 
Eq. (2). 

The deflection sought is, of course: 

y\ — K =y* 

If the new heights, y t and y 2 are given, the variation of 
length, v, will be at once given by Eq. (5). 














SUSPENSION CANTI-LE VERS. 


. 28 7 

If heights of towers are the same, Eq. (6) will not be 
needed ; for making x x = c x — c 2 , and yi=-y % = h, there 

results: 


v — 


2/ 2 
16k 



\6h 2 . t 

- /3 - + hyp. log 



— 2C X . 



In Eq. (7) h is then to be found by trial, as before, if v is 
given ; or if h is given, v at once results. 

The deflection of the middle point of the truss will be : 



It is to be noticed that in Eqs. (5) and (7) all the quantities, 
y l7 y 2 , and h, increase in the same direction with v. This ma¬ 
terially simplifies the approximation by trial. 

The determination of v in Eq. (5) might be made without 
assuming and x 2 to remain constant, for there are two other 
equations of condition: 

X\ 4 - ^2 — A 

and 


y\ y 2 * 

These, with Eqs. (5) and (6) would be sufficient in order to 
find the four unknown quantities, y u y 2 , x^ and x^- 

Such a degree of extreme accuracy, however, is unneces¬ 
sary. 


Art. 60.—Suspension Canti-Levers. 

In the figure, ABD represents a suspension canti-lever. The 
cable BC goes over either to another span or to an anchorage, 
while A is the end of the canti-lever. The cable AB is in 
precisely the same condition as the half of a cable belonging 









288 . 


SUSPENSION BRIDGES. 


to a span equal to 2 AD; consequently its tension R at any 
point and its inclination at the same point are to be found 


by the formulae already 
given. In fact, all the 
circumstances are pre¬ 
cisely the same except 



^ this, the platform is 
subjected to a thrust, 
uniform throughout its 
whole length, and equal 
to the constant hori- 


D 


Fig. i. 


zontal component, H , of the tension R. 

Art. 61.—Suspension Bridge with Inclined Suspension Rods—Inclination 
of Cable to a Horizontal Line—Cable Tension—Direct Stress on Plat¬ 
form—Length of Cable. 

In this case the suspension rods, or suspenders, are all sup¬ 
posed to be equally inclined to a vertical or horizontal line, 
and, consequently, are parallel to each other ; they are also 
supposed to take hold of the platform or stiffening truss at 
points equidistant from each other. These conditions cause 
the cable to be subjected, in each of its parts, to the action 
of parallel loading of uniform intensity in reference to the 
span. As was shown in the first Article of this Chapter, the 
curve of the cable will be composed of common parabolas 
having axes parallel to the suspension rods. 

In the figure, A is the lowest point of the cable, while BD 
and FG represent suspension rods on either side of A. The 



c 


Fig. i. 


angle a is the common inclination of all suspenders to a ver- 













INCLINED SUSPENSION RODS. 289 

tical line, it also represents the inclination of the axis of 
either of the parabolas AC or AE to the same line. 

The vertex of the parabola AC is on the left of A f and the 
vertex of AE is on the right of the same point. 

Let x be horizontal in direction and measured from A, and 
let the length of any suspender, as BD be denoted byy, but 
let CP be designated by y v Either parabola, as A C, will then 
be referred to oblique co-ordinates in the usual manner. 

If OB be drawn tangent to the curve at B, A O will be 

equal to OD , or —. If i represents the inclination of the curve 

at any point, as B, to a horizontal line, the triangle OBD will 
give: 

BD _ 2y _ sin i _ sin i 

OD x cos (a: + i) ~ cos a cos i — sin a sin V 


2 y 


cos a 


tan i — 


2y 

I + —— sm a 


. . . (i). 


In the usual manner, sec i = V 1 + tan 2 i, or, 


. V 

sec 1 — ~ 


4 y Ay 2 

I + stn <* + 


;tr 


X“ 


2y 

I -|-— sin a 

x 


. . (2). 


At the point C, if CM — h h AP= x h and AM = a, in Eqs. 
(1) and (2), /q sec a is to be put for y, and (a — h x tan tf) for^r. 

Exactly similar equations apply to the other portion of 
the span. 

For the point A, Eqs. (1) and (2) apparently become inde¬ 
terminate, but only apparently, for the relation, 

y h x seca 

?■"(*- K tan af ’ * 


19 















290 


SUSPENSION BRIDGES. 


gives, 



2 h x sec a 

(a — h x tan af X ’ 


2y 

and when x — o, consequently, — becomes zero, making tan i 
equal to zero also. 

If OBD be taken as a triangle of forces, OB will be the 
cable tension R at B ; while OD will be the horizontal com¬ 
ponent H , and BD will represent wx see a. w is the total 
load per unit of span on AD. 

From the triangle in question, 


H _ cos (a 4 - i) _ x 
wx sec a sin i 2 \y 


H = 


zvx * sec a 


2 y 


jrj. _ w(a — h t tan a ) 2 

2 ^ 



As was to be expected, Eq. (4) shows H to be a constant 
quantity, but it is not a rectangular component in this case. 

The same triangle gives for the resultant tension at any 
point: 

R = \/(wx sec of + H 2 + 2 wx H tan a. . . (5). 


For the point C, becomes (a — h x tan a). 

If / is the span, these equations apply to the other portion 
of it, by taking h 2 for k u and (/ — a) for a. 

If the towers are of equal heights, h x becomes equal to k 2 , 

and a — l — a = —. 

2 

Let p be the horizontal distance between any two suspend¬ 
ers, then the tension, t, in the suspender will be: 


t — wp sec a 



The direct stress in the platform is caused by the horizon¬ 
tal component of the tension in the suspension rods. This 
stress may exist as tension in the platform, in which case it 










INCLINED SUSPENSION RODS. 


29I 


will exert no action on the towers. Remembering that all 
the suspension rods must, at any instant, be subjected to a 
uniform stress, it is evident that the direct tension in the plat¬ 
form will have its greatest value at the centre, and will be 
equal to 

nt sin a = nwp tan a ; 

in which n is the number of suspension rods in each half of 
the span, supposing towers to be of equal heights. If n! be 
the number of suspenders between the end of the span and 
any point, the tension in the platform at that point will be 

n!t sin a = n'wp tan a. 

If the towers are of unequal heights, there will be a greater 
number of suspenders on one side of the lowest point of the 
cable than on the other. Let n x be the number in that por¬ 
tion of the span adjacent to the highest tower, and n 2 the 
number in the other portion; n x will be greater than n 2 . In 
this case, then, the platform at the foot of the highest tower 
will sustain a thrust given by the expression 

(;q — 7t 2 ) t sin a = (?i t — 71 2 ) wp tazi a. 

If the platform is to sustain a direct thrust only, at the feet 
of the two towers it will have to sustain thrusts given by the 
expressions 

n x t sin a = 7 i x wp ta 7 i a 

7i 2 t sin a — ;/ 2 wp tan az. 

If n’ represents the number of suspension rods between the 
centre and any point, the thrust at that point will be 

n!t sin a = nwp tan a. 

In the case of a suspension cantilever, in addition to the 
thrust given above there will be one denoted by H y uniform 
throughout its length. Other calculations for a suspension 
cantilever are precisely the same as those already given. 


292 


SUSPENSION BRIDGES. 


The length of the cable from the lowest point to any other 
point at which the inclination to a horizontal is i, is readily 
found by means of the formula used for the cable with verti¬ 
cal rods. In the present case the inclination of the cable at 
any point to a line perpendicular to the axis of the parabola 
is (i+a ); consequently there is simply to be found the length 
of the parabolic arc between the points at which the incli¬ 
nations to the axis are (90 — (i+a)) and (90 — a). 

The formula mentioned then gives 



(7 + a) sec (i + a) — tan a sec a + 

. . tan (i +a)+sec (i +a) 

hyp. log. ——------- 

tan a + sec a 



It is known from analytical geometry that p takes the fol¬ 
lowing form in terms of the oblique co-ordinates used in this 
case: 

_ x 2 cos 1 a _ (a — h x tan a) 2 cos 3 a 
^ 2y 2/q 


Eq. (7) is, of course, to be applied to both branches of the 
curve to obtain the total length. 

From what was said in the demonstration of the approxi¬ 
mate formula, it may be seen that it can be applied to the 
present case by changing x to (x + y sin a) and, y to y cos a. 
The formula then becomes: 


c — x + y sin a 4- 


2J/ 2 cos 2 a 
3 (x + y sin a) 



Art. 62.—Suspension Rods; Lengths, and Stresses. 

In the following calculations it is virtually assumed that the 
cable lies in a vertical plane, and that the suspension rods are 
vertical. This, however, does not affect the generality of the 
results obtained, for in all cases the suspension rods are sup¬ 
posed parallel to each other, and the lengths found by the 






SUSPENSION PODS. 


293 


formulae of this Article are to be taken as the vertical pro¬ 
jections of the true or actual lengths. The true lengths are 
therefore to be found by multiplying the values of k 0f h u 
etc., by the secant of the common inclination to a vertical 
line, of the suspension rods. 

Since a flat parabola nearly coincides with a circle, the 
camber may be supposed to be formed by a parabolic arc. 
Let the co-ordinate x be measured from A toward B , in Fig. 
14, PI. XII., and y perpendicular to it; also let AB — x Y = 
half span. Then since the curve of the cable is supposed to 
be a parabola in a vertical plane : 


y =y 1 


X d 


2 * 


In the same manner for the camber: 


y 


11 


6 


x* 


*1 


2 • 


Then the total length of any suspender is: 

h =/ + y" 


c. 


When the suspenders are separated by a constant distance, 
d, simpler formulae may be found. 

Each suspender is composed of the sum of two variable 
lengths (y and y") and a constant length, c. Now if (y L + S) 
be written for : 


y = 0'i + O 


X 2 


X 


2 > 


will evidently be the sum of the two variable lengths referred 
to. Hence, if k 0 , h lt h«, h s * * * K represent the lengths 

of the suspenders as shown in the figure: 


JlQ — c , 


d' 


h\ — c + 2 (Ti T d), 

x i 





294 


SUSPENSION BRIDGES. 


K — c + (yi + ^)> 

x i 

h = c + -^r (J'l + <*)> 

■*1 

♦ 

(« — l) 2 <^ 2 

hn-1 — c H- E2 ( y 1 + 

«*1 

h n — c + = (yi + = c + yi + 

Having computed the lengths of the suspenders for one 
half the span, the results may be used for the other half if the 
piers are of the same height ; otherwise the lengths must be 
computed separately. 

The stress in any suspension rod is the vertical load which 
it carries, multiplied by the secant of its inclination to a 
vertical line. 

Art. 63.—Pressure on the Tower—Stability of the Latter—Anchorage. 

Let P v = vertical component of pressure on tower head. 

“ P h — horizontal “ “ “ “ “ “ 

“ R = resultant “ “ “ “ 

“ T p and T p ', Fig. 13, PI. XII., = tensions of the cable on 
different sides of the pier head. 

a, a, and 6 represent inclinations to the vertical as shown. 

When friction on the saddle is considered: 

P v — T p cos a + Tp cos a ; P h = T p sin a — T p sin a '; 

R - VP v 2 + Th ; cos 6 = ~. 

When friction on the saddle is not considered, T p = T p ’; 

P v = Tp (cos a + cos a '); P h — T p (sin a — 

R = VPv + Pn ; cos 6 . 

K 


sin a') . (1). 







PRESSURE ON THE TOWER. 


295 


In the same case if a = a' ; 

P v — 2 W, P h — o, R — 2 W, 6 = o. ( W = \ weight of load 
and structure.) 

There are two cases in which the resultant pressure on the 
tower, caused by the tension in the cables, may be vertical in 
direction. Both, however, are founded *on the single condi¬ 
tion that the horizontal components of the cable tension, on 
each side of the tower head are equal to each other. 

This condition will exist if a = a in Eq. (1), making P h — o; 
or if the saddle be supported on rollers and roller friction be 
omitted. In the latter case P h — o because T p sin a = 
Tp sin a', and not because a necessarily equals a . 

In discussing the stability of position of masonry towers, 
let the distance of the centre of pressure from the centre of 
figure of the section of the pier be denoted by q. If this 
latter does not exceed q (the limit of safety for q’), which 
may be ascertained by determining the line of resistance for 
the pier, stability of position will be secured. 

It is supposed, of course, that 6 has some value greater than 
zero ; otherwise q — o. 

The stability of friction for masonry towers will be secured, 
at any joint, if the obliquity of the resultant pressure be less 
than the angle of repose. 

Iron and timber towers are to be treated, each as a whole, 
as long columns, by Gordon’s formula. 

If the anchorage is a mass of masonry, the stabilities of po¬ 
sition and friction are to be considered. 

Let W — weight of mass and q the normal distance from a 
vertical line through its centre of gravity to the centre of fig¬ 
ure of its base. Let T p equal the tension in anchor chains 
and p the normal distance from the centre of pressure to its 
line of direction; and q the distance from the centre of press¬ 
ure of the base of the foundation to its centre of figure. 

Then, in order that stability of position may be secured : 

T p p A W{q+q) 

Stability of friction is secured if the greatest obliquity of 


SUSPENSION BRIDGES. 


296 

the resultant pressure on any section (including the base) is 
less than the angle of repose for the surfaces in contact. 

If it were not for friction between the anchor chain and it's 
supports, on the circular part of the chain (see Fig. 15, PI. 
XII.), the tension would be the same throughout its whole 
length ; but on account of friction, the tension diminishes on 
the circular part, from link to link downward, according to the 
law of friction between cords and cylinders, and is, therefore, 
the least at the bottom. 

The diminution of tension of the anchor chain is computed 
by this formula: 

T p = T p E fQ 

in which T p = tension of anchor chain before friction takes 
effect; T p — tension of any point below the first point of sup¬ 
port ; E — base of Napierian system of logarithms ; f — coef¬ 
ficient of friction ; 6 =■ length of arc considered. The anchor¬ 
age can be ruptured only by the breaking of chain, or bolt, or 
plate, or pulling out the whole masonry. The probability of 
the latter can be determined by comparing the tension of the 
chain, at the upper surface of the masonry, with the weight 
of the whole masonry. 

Art. 64.*—Theory of the Stiffening Truss—Ends Anchored—Continuous 

Load—Single Weight. 

It has been seen that when a suspension bridge cable 
carries a load covering the entire span, of uniform intensity 



per horizontal unit, its centre line forms a parabolic curve. 
When, however, such a cable carries an isolated weight, or a 
partially uniform load, it is evident that the centre line of the 


* This and the following two Arts, form the substance of a paper presented 
to the Pi Eta Scientific Society, in June, 1879 , 














THEORY OF THE STIFFENING TRUSS. 


29 7 


cable will assume a form different from that of the preceding 
parabola, unless such a change is prevented by some special 
device. Such a special device is the stiffening truss. 

The objections to a change of form in the cable of a sus¬ 
pension bridge are of two kinds. Not only would destructive 
undulations result, but, also, the determination of stresses 
would become exceedingly complicated and uncertain. 

Two cases may arise: the stiffening truss may be securely 
anchored at its ends; or its ends may simply rest upon sup¬ 
ports and be free to rise, in which case there can be no nega¬ 
tive or downward reaction. 

The former case will be considered first. 

It is desired to have the cable retain, for all positions of the 
moving load, the same parabolic form. Now, it has already 
been seen that such a result can be attained only by assuming 
a uniform pull on the suspension rods from end to end of the 
span. Let T be the general expression for this uniform pull 
for any suspension rod, and let t be its intensity per unit of 
span, so that if p be the panel length of the stiffening truss, 
T — pt. Let w be the weight per unit of span of the fixed load 
sustained by the cables. This will, of course, be composed of 
the weights of the truss, suspension rods and cable or cables. 
Let w' be the moving load per unit of span ; / the span ; R 
the reaction at B ; R' the reaction at A, and let the moving 
load pass on the bridge from B. 

Also let x x be the distance from B to the head of the mov¬ 
ing load; the latter being supposed continuous from B. 

Since all the forces acting on the stiffening truss are vertical 
in direction, there are only two general conditional equations 
of equilibrium, and those simply indicate that the sum of all 
the external vertical forces, as well as the sum of the moments 
of the same, about any point, must be zero. 

Those two equations are the following: 

wl + w'x t —tl — R —R' = o . . . . (1). 




R' {l—x j) = o. . . {2). 


2 





SUSPENSION BRIDGES . 


298 


Eq. (2) can be at once written by taking moments about 
the point x x at the head of the moving load. 

Eqs. (1) and (2) are the only equations of condition neces¬ 
sary for equilibrium, but they hold three unknown quantities, 
i. e., t , R, and R' ; hence, any one of those three quantities 
may be assumed at pleasure, and the other two determined 
from Eqs. (1) and (2). This indetermination simply means 
that unless another condition be imposed, it cannot be ascer¬ 
tained how much the truss will carry as a simple truss , and 
how much in connection with the cable. 

This other condition is virtually the following: the stiffening 
truss must act wholly in connection with the cable , and carry 710 
load whatever as an ordinary truss . 

The direct consequence from this condition of the problem 
is, that the sum of all the uniform upward forces, T = //, 
must be equal in amount to the sum of all the loads of the 
kinds w and w. But the line of action of the resultant of the 
latter is not, for a partial moving load, the line of action of 
the resultant of the former: consequently, the truss will be 
subjected to the action of a couple. In order that equilib¬ 
rium may be assured, therefore, another couple of equal mo¬ 
ment, but opposite sign, must be applied to the truss; the 
forces of this couple must act at the extremities^ and B, and 
they are nothing more than the reactions R and R\ From 
this there at once results: 


R= -R'. 


This condition, in Eq. (1), gives: 




Eq. (2) gives: 




Eq. (4) shows that both reactions, R and R r , are zero for 
x x = /, or for w' = o. 




THEORY OF THE STIFFENING TRUSS. 2 gg 

It is also seen that R and R' are always numerically equal, 
but have opposite directions, hence R' is a downward reaction, 
and its maximum value will indicate the amount of anchorage 
required at each end of the truss. 

Using Eq. (4): 


dR w' w'x-, w'x t / 

~r - =-- f -= O, X X = 

dx x 22/2/ 2 


Putting x x = — in Eq. (4) : 
2 




Eq. (5) shows the maximum value of (— R f ) and gives the 
amount of anchorage required at either end of the truss; it 
aiso shows the greatest shear to be provided for at either end 
of the truss. 

The general value for the shear at any section of the por¬ 
tion of the truss covered by the moving load is: 



S = R + tx — wx — w’x .... (6). 


r 



W'x i 2 / Xi 

ST + X \T 



• This value of 5 shows it to be positive near the end of the 
bridge ; it then decreases as x increases, passes through the 
value zero, and then increases as a negative quantity. As a 
negative quantity it attains its maximum value for x — x x ; it 
then becomes: 






Hence the tzvo reactions at the ends of the truss and the shear 
at the head of the moving load are always numerically equal. 









300 


SUSPENSION BRIDGES. 


Eq. (8), consequently, takes its maximum value for x 1 = —, and 

that value is given in Eq. (5) ; this last equation, therefore, 
gives the maximum shear which is to be provided for at the 
head of the moving load. 

Now since this maximum shear is to be provided for at 
both ends and at the middle of the truss, it would probably 
be advisable in all ordinary cases to design all the web 
members of the truss to be of uniform size, and capable of 
carrying this maximum shear, although there would then be 
a little waste of material in the vicinity of the quarter points 
of the span on each side of the centre. 

This supposes, of course, that the chords of the stiffening 
truss are parallel and horizontal. If the chords are not 
parallel the amount of shear carried by the web members 
will depend on the inclination of one or both the chords. 

For all values of x and x 1} for the portion of the span cov¬ 
ered by the moving load, the total shear will be given by 
Eq. (7). 

For the portion of the span not covered by the moving 
load the general value for the shear is (measuring x from A) : 



R' + wx — tx = 


w'x\ 

2 


f 2 

W Xi 

21 


w'x^x 

~~r 



This expression attains its greatest values for x = o and 
x — l — x v In the first case the shear becomes — R\ and 
in the second R. These results show nothing new. 

Since the maximum shear in a simple truss, of the span / 
and uniform loading of intensity (w + w'), is \{w + zv')l, it is 
seen that the maximum shear in the stiffening truss of same 
span is only one-fourth of that due to the moving load alone 
in the case of the simple truss. 

The general value of the bending moment to which the 
truss is subjected, for the portion covered by the moving 
load, is: 

* 

M = Rx — (w + w f — /) — . . . (10). 

2 





THEORY OF THE STIFFENING TRUSS. 


301 


Eq. (10) shows that \i x = x\, the bending moment is equal 
to zero. Hence, at the head of the moving load, for all its 
positions, there is a section of contraflexure or no bending, and 
consequently the loaded and unloaded portions of the stiffening 
truss are each m the condition of a simple beam supported only 
at each end, and loaded uniformly throughout its length. 

If the values of R and t from Eqs. (3) and (4) be inserted 

in Eq. (10), and if -- be put equal to zero, it will be found 

IT 

that the bending moment has its maximum value for x — 

2 

as might have been anticipated. 

Putting .r = — in Eq. (10) : 

2 


M = 


w' x 2 
8 



(11). 


By differentiating in respect to x x it will be found that M 
has its maximum value for x x = §/. Denoting this value of 
M by M lf there results : 


I 



(12). 


Eq. (12) shows the maximum bending moment to which 
any loaded portion of the truss can be subjected. 

If a simple truss supported at each end be subjected to the 
action of a uniform load, of the intensity {w + w'), through¬ 
out its entire length, the greatest bending moment will be: 


M' = 


(w 4- w') l 2 _ w'P 


8 


8 


(if w = 0); 


Mi = A Mi > \ Mi 


( 13 )- 


* 


From what has already been shown it is evident that the 


* The subscript o indicates that w = o in M'. 









302 


SUSPENSION BRIDGES . 


greatest bending moment for the portion of the truss not 

covered by the moving load will occur at the distance ^ ^ 

from the reaction R. The general value, therefore, for the 
greatest moment for that portion will be : 


M=lR'(l-x l ) + (t- w) f - . . (14)- 

Putting *the differential coefficient of M in respect to X\ 
(after inserting the values of R and t) equal to zero, there 
results : 


*i = f/± - |/ 2 

x x — \L± — — lox\l . . . . (15)- 

3 

The latter value (1/) gives a maximum, and inserted in 
Eq. (14): 

= - — - - Mi. = - IM 0 ' (nearly) . . (16). 

54 

Eqs. (12) and (16) show that the greatest bending moments , 
to which the stiffening truss is subjected , are equal , but of oppo¬ 
site kinds; and it is seen that they occur zvhen one-third or 
two-thirds of the span are covered by the moving load. The 
chords, therefore, of the stiffening truss must be designed to 

resist both tension and compression. 

Eqs. (10) and (14) are general expressions for all the bend¬ 
ing moments to which any portion of the truss can possibly 
be subjected, but in all ordinary cases it would probably be 
best to make the chords uniform in section (supposing the 
depth of the truss to be constant) from end to end, and 
capable of resisting the moments given by Eqs. (12) and (16). 

Eq. (13) shows that the greatest bending moment to which 
a stiffening truss can be subjected is only ^ of that found in 
a simple truss supported at each end and loaded with a uni- 






THEORY OF THE STIFFENING TRUSS. 


303 


form load equal in intensity to that of the moving load on 
the stiffening truss. 

It may be interesting to notice that the resultant load on 
the portion x x of the truss is downzvard, since (w + zv') > 


(^t = w + 


zv'x A 

) 


but that that on the portion (/ — x x ) is 


upward, since zv < t. 

If the bridge is traversed by a single concentrated load or 
weight, IV, the general method of procedure is precisely the 
same as before. Let the weight W pass on the bridge from 
the end B, in the figure, and let x x denote its distance from 
that point; also measure x from the same point. 

The general equations of condition are: 


R + R + tl — zvl — W= o . . . (17). 

X ,) + (w - t) I—Nt _ _ t) -d + R Xl = o . (18). 


Eq. (18) is written at once by taking moments about the 
point of application of W. 

By precisely the same method as before, there may be found 
the result, R = — R '. 

In Eqs. (17) and (18) let there be put R = — R' y then there 
results: 


W 

t = w+ T 



R=-R'= . . . (20). 

When x x — —, R — o and R — o 

Eq. (20) shows that the reaction nearest the weight W will 
always be positive, or upward; and that the other will be nega¬ 
tive, or downward. 





304 


SUSPENSION BRIDGES. 


The maximum value of R or R' is found (by making x 1 — o 

W 

in Eq. (20)) to be —, and that is the amount of anchorage 

required for the weight W, alone, at each end of the truss. 

The point of application of the weight W divides the span 
intp two segments. 

The general value of the shear for the shorter segment is: 

S = tx -wx + R = ~x + wU -yj . . (21). 
This has its greatest value for x = x x ; it then becomes : 



Hence S' is the uniform maximum upward shear that must 
be provided for , throughout the whole length of the truss. 

The general value of the shear in the longer segment of 
the span is : 

S=R' - w(l-x) +t(l-x) = w(^ + y -j\ . . (22). 

i 

This expression attains a positive maximum for ;r = x 1} 
that being the least positive value of ^ admissible ; the re¬ 
sulting value of S is \ IV, which shows nothing new. 

The negative maximum for x — l is simply the reaction R\ 
Putting 5 = o in Eq. (22) there results : 

/ 

x = Xi -\— . 

2 

Hence for all points of the span between — + x x and / there 

will be negative or downward shear. The maximum negative 
shear, however, to be provided for, is shown by Eq. (22) to 
exist in one-half of the truss when W rests on the opposite end; 

or, when in that equation x x = o and x > — . The negative 

2 


THEORY OF THE STIFFENING TRUSS. 


305 

or downward shear to be provided for has, then, for its gen¬ 
eral expression : 


•Si = - W (2 - j) . . . . (23). 

Eq. (23) is to be applied to each half of the truss, and it is 
also seen that the web members which take up S 1 should 
increase, in ultimate resistance, uniformly from the centre to 
the ends of the truss; supposing the chords to be parallel 
and horizontal. 

In many cases, however, it may be best to design them of 
uniform dimensions belonging to those at the ends. 

The bending moment for any point of the smaller segment 
of the truss is : 


M=Rx + (, t-w )y= w(^-j)x + 


Wx * 
2 1 


. (24). 


Since both terms of this moment are positive, it will attain 
its greatest value for x = x x ; it then becomes : 


^ w / x?\ , x 

M \ — ^ (^1 ^ J • • • ( 2 5)- 


Eq. (25) gives the general value of the greatest positive 
bending moment to which any point of the truss will be sub¬ 
jected, for which case x x must never be made greater than \ l. 

The bending moment M l will evidently cause compression 
in the upper chord, and tension in the lower. 

For the longer segment of the truss the general value of the 
bending moment is: 

M — R' {l — x) + {t -w f ~ . 





(l — x) + 


W (/ -xf 
~T 2 


20 


. (26). 






SUSPENSION BRIDGES. 



There is evidently a point of contra-flexure for the longer 
segment of the truss, for if the second member of Eq. (26) be 
put equal to zero, there results — 2 x v Hence all the por¬ 
tion l — 2x x , of the truss , zvill be subjected to a negative bending 
moment causing tension in the upper chord and compression in 
the lower . 


In Eq. (26), putting 


dM 

d (l — x) 


= o, there results : 


x = x^ 




This value of x, in Eq. (26) gives : 


M' 


Wl , Wx x Wxf 
8^2 2/ 



Eq. (28) gives the general value for the maximum negative 
bending moment at any point in the entire truss , for any position 
of the weight W. In Eq. (28), it is to be remembered, x x must 

l 

always be less than —; also, that the point at which M' exists 

will be given by the value of ;r in Eq. (27). 

The formulae for a continuous load, taken in connection 
with those for a single weight, will give all the circumstances 
of bending or shearing which can exist with any condition or 
position of loading. 

When the “ shear ” has been determined for any section, 
the stress in the web member which is to carry it (if the 
chords are parallel and horizontal) will be found at once by 
multiplying that “ shear” by the secant of inclination of the 
web member to a vertical line. If there are two or more 
systems of triangulation in the truss, then each system is to 
be treated as a single truss in the usual manner. 

If desirable, after the reactions R and R' and the upward 
load T — pt are known, the stresses in the individual members 
of the stiffening truss can be traced as in the case of an ordi¬ 
nary truss supported at each end. 







THEORY OF THE STIFFENING TRUSS. 


30 7 


Art. 65.—Theory of the Stiffening Truss—Ends Free—Continuous Load 

—Single Weight. 

In this Article the notation of the previous one will be con¬ 
tinued, and the same figure will be referred to. 

The case of a continuous load will first be treated, and, as 
before, it will be supposed to pass on the bridge from B. 

Since the ends are not anchored, in' this case there can be 
no negative or downward reaction, consequently R' will be 
zero. 

As before, putting the sum of all the vertical forces acting 
on the truss equal to zero, and taking moments about the 
head of the moving load, there result the two general Equa¬ 
tions of condition : 


wl -f w x x — tl — R — o .(1). 


(w + w — t) — —(w 
2 


t) ^ - Rx x = o . . (2). 


Since there are now but two unknown quantities, R and /, 
the problem is perfectly determinate. Eqs. (1) and (2) give : 


r = w'x 1 (i- x y 


(3) 


t = w + 


w'x x 

~l*~ 


. . . (4) 


The general value for the shear at any section of the truss, 
for the portion covered by the moving load, is 

5 = R + {t — w — w) x = w (pc x — x) — w I — -j^J • ( 5 )- 


Evidently S' has its maximum positive value for x — o; its 
greatest negative value for x == x h and the value zero for x = 
lx x 

T+x; 


t 








SUSPENSION BRIDGES. 


303 

In order to find the head of the moving load for that po¬ 
sition which makes R a maximum, let —j— be put equal to 


dx-^ 


zero. 


There is then found x x — —. Hence the moving load cover¬ 
ing half the span gives the maximum reactioyi R. 

I 

Placing x x — — in Eq. (3), the greatest value of R becomes 
Rx = w' l -f- 4. 

In order to determine the greatest web stresses it is neces¬ 
sary to find the greatest shear at any point whose abscissa is 
x . This maximum shear at once results by placing the first 
derivative of 5 in respect to x lt from Eq. (5), equal to zero. 
That operation gives: 

2 X X 


1 — 


x \ / 2 

1 -7) = O .*. X x = —rj -r . 

1 ) 2(1 - X ) 


By the introduction of this value of x l in Eq. (5), the great¬ 
est shear for any section located by becomes: 

1 / 2 ) 


max. S = w 1 


x 


4(1 - x) 

It is clear that this value is a maximum for the reason that 
d> S 


dxi 


is a negative quantity in which x x does not appear. 


It is further evident that max. S is a positive, or upward 
shear, from the fact which was observed at the bottom of 
the preceding page, that the greatest negative shear occurs 
at the head of the moving load. By making x = in Eq. 
(5) that greatest negative shear becomes : 

2 / 


Si = - v/ 


xp 

l 


1 ~T. 


It will be necessary to apply max. S and to a half of the 
span, regarding the shear for a positive direction on one side of 
the centre as negative for the other. These two values of the 
shear will enable all the greatestweb stresses to be determined. 

Since R' — o, the whole truss will be subjected to bending 
moments of the same sign ; such bending moments, in fact, as 







THEORY OF THE STIFFENING TRUSS. 309 

will put the upper chord in compression and the lower one in 
tension. 

The general value of the bending moment for that portion 
of the truss covered by the moving load is. 


M — Rx — (w -f w' — t) 


x 2 

J’ 


IX/r , w'x?x w'x 2 w'x^x 2 

M = w x x x - - ---1- 7 \.— . . 


/ 


2/ 2 


( 6 ). 


Since the shear S' is zero for x = 


lx x 


l 4 - x t 


, that value of s in 


Eq. (6) will give the maximum value of M. This latter is : 


_ / 2 ^ ^1 


Mi = w x 1 


2 (l + Xi) 


■ ( 7 )- 


^ . dM 

rutting — 
dxi 


o, there is found, 


X\ — -(—1 ± Vs) — + 0.618/. 

2 

The absolute maximum bending moment exists, therefore, 
when the moving load covers 0.618 of the span. That mo¬ 
ment has for its value : 

0.045 1 w 'l 2 (nearly).(8). 

lx 

If x x = 0.618/be put in the expressions = — ' 1 —, there will 

/ + Xi 

result s = 0.382/, nearly. 

Eq. (6) gives the general value of the bending moment for 
any position of the load, but it would probably be the most 
convenient to make the moving load cover 0.618/, and design 
the chords for the distance 0.382/ from each end to resist the 
bending due to that position of the load, and then design all 
of the middle 2 (0.5 —0.382)/= 0.236/ to resist the moment 










510 


SUSPENSION BRIDGES. 


given by the expression (8). By this arrangement there would 
be a little surplus of material at the middle of the truss. 

If a single weight rests upon the bridge, the two general 
equations of equilibrium, obtained in precisely the same man¬ 
ner as heretofore, are : 

wl + W — tl — R — o .(9). 

(w — t ) — — (w — t) ——— 1 ^- — Rx x = o . . (10). 

2 2 

These equations then give: 


t — w + 


2 Wx 1 

l 2 



R= IV(l ...... ( 12 ). 

T t l W 

If x t = —, t = w + —, and R = o. 

The general values of the shear S, and moment M, are the 
following: 

= R + (t — w) x, 


\ S ~ w( i - + 2 






• ( 13 )- 


M = -f (/ — w) 


4 T 2 


\ M= Wl 1 - 


2,t"i \ _ t 


/ 


* + 


P 


• • (14)- 


Eqs. (13) and (14) show, since x 1 must not be taken less 
than that if the maximum shear and bending moment are 
desired for any section, the weight, W, must be placed at that 
section, and x be made x x in those equations. 









APPROXIMATE CHARACTER OP INVESTIGATIONS. 

That section at which the bending moment will attain its 
absolute maximum value is found by putting ;r=;r r in Eq. 
(14), then taking the first differential coefficient of M in re¬ 
spect to x 1} equating to zero and solving. There results: 

x \ — \l ± \l — 

This value in Eq. (14), when x = x h gives: 

M= ftWl. 

The equations for the continuous and single moving loads, 
used in combination, will give moments and shears for any 
character and position of loading whatever. 

The general observations in regard to finding web and 
chord stresses, at the close of the last Article, apply equally 
well to this case. 

As was to be anticipated, R and R' in the two preceding Ar¬ 
ticles, are independent of the fixed load w. 


Art. 66.—Approximate Character of the Preceding Investigations—De¬ 
flection of the Truss. 

In the two preceding Articles it has been virtually assumed 
that the deflection of the cable, due to its lengthening under 
stress, is just sufficient to allow the truss to take the deflec¬ 
tion due to the loads T — pt, w, and W in the different cases. 

Such, however, is not really the case. 

In all ordinary cases, the cable does not deflect to that ex¬ 
tent. The result is, as is evident, that the truss is not sub¬ 
jected to the amount of bending assumed. The error, 
however, is only a small one, and on the safe side, as should 
be the case. 

It has been found by experiment that if the ends of the 
truss are anchored, the stiffening truss will be subjected to a 
maximum moment equal to that existing in an ordinary truss 
supported at each end, of about one-eighth the span, and 


312 


SUSPENSION BRIDGES . 


carrying load over its entire length of the same amount as 
that of the moving load on the suspension bridge. 

This same case, treated analytically, as was seen, gave about 
\ instead of J. 

Approximate values of the deflection of the stiffening 
truss can be found by the ordinary formulae used for solid 
beams in the subject of resistance of materials, in the differ¬ 
ent cases, where t , R , R', and the moving load are known. 


CHAPTER X. 


DETAILS OF CONSTRUCTION. 

Art. 67.—Classes of Bridges—Forms of Compression Members—Chords 

Continuous or Non-continuous. 

Regarding the systems of construction, truss-bridge struct¬ 
ures are divided into two classes at the present time, i. e ., 
bridges with “ pin connections ” and bridges with “ riveted 
connections.” In the former class the connection of web 
members with the chords or with each other is made by a 
single pin only, as in Fig. 3, PI. III. (the figure shows simply 
the upper chord and tension web members together witn one 
end post). The pins are shown at 1,2, 3, 4, and 5, where the 
tension members join the chord. In the latter class the con¬ 
nections mentioned are made by means of rivets, as shown in 
Fig. 6, PI. XI. In that figure A is a tension, and B a com¬ 
pression web member, while C is a portion of the lower 
chord. 

Screw connections for tension web members and simple 
abutting connections for compression ends have been used, 
but are not usually employed at present. 

A screw connection is formed by passing a tension member 
through the chord at one of the joints, and placing upon the 
end of it a nut; this method, therefore, can only be conveni¬ 
ently used when the tension members are of circular cross- 
section. 

An abutting connection for a compression member is 
formed by simply abutting either end against the chord, 
which is properly formed for the purpose at the joint. The 
end of the post or strut is inserted in the chord, or else a 
projection of the chord passes into the end of the post; or, 
again, some simple device is employed for the purpose of 

313 


314 


DETAILS OF CONSTRUCTION. 


keeping the ends of the post in position, and for nothing else, 
as the entire compressive stress in the post or strut is trans¬ 
mitted through the abutting surfaces. 

Occasionally screw, abutting, and pin connections are com¬ 
bined in a single bridge. 

Without regarding systems of construction, truss-bridges 
are divided into: 

“ Deck ” bridges, i. e. y the applied load is on the chord in 
compression. 

“ Through ” bridges, i. e., the applied load is on the chord 
in tension. 

“ Pony ” trusses are through truss-bridges when the trusses 
are not sufficiently high (or deep) to need overhead cross¬ 
bracing, and they are seldom put up for spans of over eighty 
feet, although there are examples of one hundred feet (and 
even more) in length. 

The lateral stability of such long pony trusses, however, is 
very precarious. 

Figs, i and 2 of PI. XI. show the ordinary forms of plate 
girder, stringers and floor beams, with plate hangers at the 
ends of the latter. 

The various forms of cross-sections of upper chords and 
posts, and compression members generally, are almost innu¬ 
merable, and subject only to the fancy of the engineer or 
builder. The principle which should always be kept in view 
is this: That the material should be as far as possible from 
the neutral axis. Figs. 3, 4, 5, and 6 of PL III. show methods 
of building up the upper chord. It consists of riveting plates 
to a pair of channel bars, or to a pair of channel bars and an 
X beam. 

The bars or beams are frequently built of plates and 
angles. 

The blackened portions represent sections. 

Chords are continuous or non-continuous according as they 
are built up in a continuous manner from end to end, or built 
up of panels abutting against each other at the panel points. 
The former are principally used at present. 

Tension members are always of rectangular or circular sec- 


CUMULATIVE STRESSES. 


315 


tion. Fig. 5, PI. XII., is a lower chord “ eye-bar.” In writing 
of channel bars, X beams, angle-irons, iron bars and rods, 
they are indicated as follows: C, I, L> □ , O; that is, by 
skeletons of their sections. 


Art. 68.—Cumulative Stresses. 

Stresses are said to be cumulative in any part of a struct¬ 
ure when they are transmitted through that part to other 
parts, whose whole duty is to sustain them, the part in ques¬ 
tion being subject at the same time to its own stress. The 
member in which the stresses are cumulative is, therefore, 
overstrained to some extent in some one or more portions of 
it. The two channel bars in Fig. 3, PL III., are the portions 
of that upper chord which are subjected to cumulative stresses. 
If C'C' is supposed to be the centre line of the bridge, then 
the compressive stress in the chord increases as C'C' is ap¬ 
proached from the end, in consequence of the components in 
the direction of the centre line of the chord of the stresses in 
the inclined ties, A , B , C , etc. This increase of stress is pro¬ 
vided for by riveting plates to the upper flanges of the two 
£s, as shown in the figure. It is evident that the plates do 
not receive the stress which they are intended to bear, except 
indirectly through the [Is and the rivets which connect the 
latter with the plates. Now since the Cs are supposed to 
have their own share of direct compressive stress to sustain, 
it is plain that the material in the vicinity of the front of the 
pin (looking from the centre of the pin toward the centre of 
the bridge) is subjected to a much greater intensity of com¬ 
pressive stress than should exist in the structure. This re¬ 
lates only to the material in front of the pin, and that is the 
only vicinity in which cumulative stresses would exist if the 
plates could be so securely riveted to the Cs that the whole 
chord could be depended upon to act as one piece. In prac¬ 
tice, however, no such riveted work exists. The Cs must in¬ 
evitably yield to some extent before they bear sufficiently on 
the rivets to give to the plates their proper share of the 
stress. The result is that not only the material in front of 


3 i6 


DETAILS OS' CONSTRUCTION. 


the pin but the whole of the Cs are overstrained by these 
cumulative stresses. The only remedy is to so proportion the 
chord that those parts which are designed to sustain stress 
shall receive it immediately, and not indirectly through some 
other part. 

The Fig. 5, PI. III., shows a method of accomplishing this 
object. The plate ab is a light one riveted to the top flanges 
of the Cs, and extends throughout the whole length of the 
chord. The increase of the areas of cross-sections are ob¬ 
tained by riveting plates to the flat sides of the Cs, and by 
adding an X, if necessary, as shown. The parts of the chord 
thus receive stresses immediately from the pins, and cumula¬ 
tive stresses are obviated. 

It is also evident that if the stresses are applied to the cen¬ 
tres of gravity of the cross-sections, or parts of the cross-sec¬ 
tion, no cumulative stresses will exist. 

Cumulative stresses are as liable to occur in riveted connec¬ 
tions as in pin connections ; in fact, more so. It may be said 
to be impracticable to so construct riveted work that cumula¬ 
tive stresses will not exist, and in this respect pin connections 
have the advantage of riveted connections. Fig. 6, PI. XI., 
illustrates the matter for a riveted chord. The plate C is 
common to the whole chord, and all web members are riveted 
to it, as shown by A and B, so that before the rivets can take 
their share of the stress in transferring it to the plate and Ls 
ED , it (the plate C ) will necessarily yield to such an extent 
that cumulative stresses will exist throughout its whole length 
to a greater or less degree. 


Art. 69.—Direct Stress Combined with Bending in Chords. 

If direct stress is not applied to the centres of gravity of 
the ends of a piece subjected to compression, it is clear that 
bending must take place. 

In Figs. 3 and 4 of PI. III., the horizontal components of 
the oblique forces in the ties A , B } etc., do not act through 
the centres of gravity of the sections of the chord, hence 
there must be a bending in those chords. If the chords were 


DIRECT STRESS WITH BENDING IN CHORDS. 


317 


perfectly straight, and if the centres of the pin-holes were all 
at the same distance from the centres of gravity of the dif¬ 
ferent cross-sections, as well as in the same straight line, then 
the total direct stress to which the chord is subject at any 
section would produce bending at that section, and the lever- 
arm would be the same for all sections. Camber and deflec¬ 
tion from loading, however, so complicate the matter that it 
is quite impossible to make even a satisfactory approximate 
computation of the chord bending arising from this cause. 
All chords in compression, therefore, should be so designed 
that the axes of the pins may traverse the centres of gravity 
of their sections, even though the ties rest directly on the 
upper chord. It is clear that when this bending exists, the 
proper distribution of direct stress is greatly disturbed, though 
to an indeterminate extent ; and is, except in most rare cases, 
a very faulty construction. 

If it be supposed that the total direct stress in the chord 
acts as if the latter were perfectly straight, so that it all 
produces flexure; and if it then be supposed that the in¬ 
crement only, at each pin or panel point produces flexure in 
the adjacent panel; it is evident that the first supposition 
will make the flexure the greatest possible, while the second 
will make it the least possible. 

It is farther evident that if cumulative stresses occur, flex¬ 
ure must necessarily exist, for the simple reason that the di¬ 
rect stress is not uniformly distributed over the cross-section 
of the chord. 

Although this flexure is indeterminate in amount and shows 
a faulty design, the attempt to utilize it has sometimes been 
made, and the analysis on which the practice was based will 
now be given, it being premised that the ties are supposed 
to rest directly on the chords, as shown in Figs. 4 of Pis. III. 
and I. 

Suppose, in PL III., Fig. 6 to be an enlarged cross-section 
of the chord in Fig. 4, and lety%* P ass through the centre of 
gravity of the cross-section, being parallel to ab and cd'; 
then, since the increment of the chord stress transmitted 
through the pin from the ties AA is applied to the cross-sec- 


318 


DETAILS OF CONSTRUCTION. 


tion of the chord at a distance h below the centre of gravity, 
there will be an excess over the uniform intensity of stress in 
the cross-section at the lower side EE, and a deficiency at the 

upper side ab. 

% 

This excess or deficiency (the same in amount, of course, 
for certain sections only) is found in a very simple manner, as 
follows: 

Let P be the increment of direct compressive stress given 
to the chord by the ties AA, and let P t be the total direct 
compressive stress in the section. Put 5 for the area of the 
cross-section. 

Now the variation of the intensity of stress from the mean 
is due to the moment Ph, and since this moment is constant 
for all points between any two pins, as I—2 or 2—3, Fig. 4, 
PI. III., the variation in intensity is also constant between 
these points. 


PI 


The moment Ph — , in which d x equals the distance 

"1 

from fg to EE, and R the intensity of stress at EE due to 
bending, gives 


R 


Phd\ 

~T 


The intensity of stress at ab due to bending is, of course, 
equal to 


d — d-[ 


R. 


Now the total intensity of stress at EE, Fig. 6, PI. III., is 
equal to — + R ; and that at ab, — R —^— 1 - ; the inten- 

p 

sity at fg= — , whatever may be the figure of the cross-sec¬ 
tion. The variation of intensity at any point in the section 
may be easily found from R by a simple proportion, and the 


total intensity by adding that to 


Px 

S' 






DIRECT STRESS WITH BENDING IN CHORDS. 


319 


As a first case, let the chord be a non-continuous one, so 
that each panel, so far as the panel moving load is concerned, 
is a simple beam supported at each end. 

If the load rests on the upper chord immediately, as shown 
in Fig. 4, PI. III., it will produce tension at the lower side of 
the chord and compression at the upper by simple flexure, an 
opposite tendency to that exerted by the moment Ph . 

The moments due to the moving load on any panel vary 
(that is, increase) from the joints to the middle point of the 
panel, where, of course, the moment is maximum. Denote 
by R! the greatest intensity of the tensile stress caused by 
the moment of the moving load, then 

D , dxSwxx 
R =—7—, 

in which 2 zux x expresses the greatest moment of the applied 
load. 

For a uniform load : 


JZwx 1 = 


wl 2 


y 


and it exists at the centre of the panel. Now, ordinarily, the 
chord would be required to resist the bending moment ex¬ 
pressed by 2 wx, but h may be so chosen that for its max¬ 
imum value R — R', and then no extra metal will be required 
on account of the flexure produced by the moving load. This 
value of h is found as follows : put 

Phd x d x 2 wx x _ 2 wx j 

— / ; tl ~ p • 


In all ordinary cases of uniform load 


2wx x = 









DETAILS OF CONSTRUCTION. 


320 

in which w is the intensity of uniform load. 

_ wP 

*’• 1 ~ ~8A * 

The value of k, therefore, is independent of the form of 
cross-sections. If R < R ', additional material will be needed 
in order to prevent an excess of compressive stress at the 
upper part of the chord, and a deficiency at the lower side. 
When R > R', there is an excess of compressive stress at EE, 
Fig. 6, PL III. 

When the lower chord sustains the moving load directly, 
as in Fig. 4, PI. XI., the only change arises from this: That 
it is in tension instead of compression, and h is measured 
above the centre of gravity of the cross-section, instead of 
below it ; also, the section is rectangular. 

In the case of the lower chord, however, if the stress in one 
panel is given to the adjacent one through the medium of a 
pin, then the total stress in the panel under consideration 
must be put for P in the formulae above. This must also be 
done in every case where the total chord stress produces 
bending. The chord stress used may be taken as the max¬ 
imum (that which exists when the moving load covers the 
whole truss), for in all other cases there is a surplus of ma¬ 
terial with which to resist the bending. 

This method of neutralizing the flexure produced by the 
direct application of the moving load to the chords is very 
unsatisfactory in many ways and should never be used. At 
all places in the panel except the centre the metal is still 
over-strained by the flexure due to its own stress, and in the 
vicinity of the panel points this condition exists to a very 
serious extent. When this consideration is coupled with the 
great uncertainty attached to the hypothesis on which the 
analysis is based, the unsatisfactory character of the method 
is sufficiently evident to effect its exclusion from the best 
practice. 

If the chord is continuous, the objections to the method al¬ 
ready mentioned gather considerably increased force. At 



DIRECT STRESS WITH BENDING IN CHORDS. 321 


and near the ends of the panels the fixed and moving load 
produce flexure in the same direction as the direct chord 
stress, and to twice the amount of that at the centre. It is 
not necessary, therefore, to consider this case farther. 

A considerable saving of material can be effected by plac¬ 
ing the ties directly on the upper chord of a deck bridge, and 
with a proper design it in no manner conflicts with the best 
practice. In all cases the axis of the pin should traverse the 
centre of gravity of the chord section , or as nearly so as practi¬ 
cable, in order, if possible, to eliminate all flexure due to the 
direct chord stress. The chord section should then be so 
formed that the combined stresses due to flexure in the ex¬ 
terior fibres, and the direct chord stress shall at no point ex¬ 
ceed a proper value per square unit. This value may be 
taken at 8,000 to 9,000 pounds per square inch for wrought- 
iron upper chords, or 10,000 to 11,000 for mild steel members 
of the same kind. These values may be taken comparatively 
high for the reason that an indefinitely small portion only of 
the material is subjected to these intensities, and that small 
portion is well supported against fatigue by the material 
about it, which is considerably understrained. 

If the notation previously used in this Art. be still main¬ 
tained, the maximum external moment 2 wx x will develop in 
the most remote fibres at the distance d x from the neutral 
axis the intensity. 


d 1 2wx l 


(0 


1 /_— •' 

/ 


If, on the other hand, P x is the total direct stress in the 
chord and 5 the area of cross section, while p is the greatest 
allowable combined stress, then there will result: 



(2). 


p = —t + 1 


Eq. (2) shows that in the most efficient design the moment 
of inertia I of the section must be the greatest possible. At 
the same time considerations affecting the joint details render 


21 






322 


DETAILS OF CONSTRUCTION. 


it advisable that the centre of gravity should lie not far from 
the mid-depth of the section. These two conditions are ful¬ 
filled by placing large quantities of the material, and as 
nearly as possible in equal amounts, at the top and bottom 
of the chords, as shown in Fig. 18 of PL XII. The cover plate 
be and angles dd are made as light as the circumstances of 
proper design will permit, but the angles aa are made as 
heavy as possible. An unequal-legged angle is a very good 
one for aa with the longest leg horizontal, and it is sometimes 
necessary to rivet a narrow plate to those horizontal legs in 
order to properly balance the section. The centre of gravity 
\\ne fg will usually lie a little above the centre of figure. 

As the centre of the span is approached from the end the 
chord section must be rapidly increased but in no case should 
that increase be made by thickening the cover plates or increas¬ 
ing their number, as such an operation inevitably means 
cumulative stresses or flexure by direct stress. In rare cases 
it may be admissible to slightly thicken a cover plate,*if there 
is but one, but, as a rule, Fig. 6 of PI. III., shows a design to 
be carefully avoided. All increase of section should be ob¬ 
tained by thickening the side or web plates, or increasing 
their number ; or, again, by increasing the angles, or, finally, 
by introducing an interior eye-beam, as shown in Fig. 5, 
PI. III. 

If the chord is non-continuous, 2 wx 1 is simply the bending 
for a span equal in length to a panel and due to the track 
load, own weight and superimposed moving load, and is easily 
determined. If the chord is continuous, on the contrary, the 
analysis for the moving load bending is not simple. The 
total bending for this case, however, may properly and safely 
be taken at three-fourths its value for a non-continuous 
chord. 

The upper chord section required in the case of combined 
bending and direct stress is readily found by the aid of Eq. 
(2). The radius of gyration r can be easily and with suf¬ 
ficient accuracy predetermined ; so that Sr 2 can be put for 1 
in that equation. After that substitution is made, there at 
once results : 


RIVETED JOINTS. 


323 



This is a very convenient formula for practical use. 


Art. 70.—Riveted Joints and Pressure on Rivets. 

In riveted bridge work the pitch of rivets (i. e., the dis¬ 
tance from centre to centre) should not be less than three 
diameters, although it sometimes is; if possible it should be 
from four to eight diameters of the rivet, provided that value 
does not exceed about fourteen or sixteen times the plate 
thickness. The diameter of the rivet is determined by the 
amount of stress which the joint is to carry, so that the in¬ 
tensity of pressure against the surface of the rivet in contact 
with the plate shall not exceed a given value. If the rivet 
and hole were in ideally perfect contact, this intensity could 
easily be found, having given the amount of stress which the 
rivet is to carry. But such is never the case. The only re¬ 
sort left, therefore, is to assume that the rivet does fit per¬ 
fectly, and fix a low enough value for the intensity of pressure 
against its surface to make the joint safe. 

In Fig. 1, PI. XII., suppose EF to be a part of a plate in 
which is drilled or punched the rivet-hole ADBK, and sup¬ 
pose the stress to be exerted on the plate in the direction of 
the arrow at K , then the surface of contact between the plate 
and rivet will be projected in ADB; contact will not take 
place throughout the whole semi-circumference when the 
plate is not subjected to stress, unless the rivet fits the hole 
with absolute accuracy. 

Since all material is elastic to some degree, there will be a 
surface of contact when the plate is subject to stress, even 
when the rivet does not accurately fit the hole, and this sur¬ 
face will evidently increase with the stress in the plate. 

But suppose that the rivet fits the hole exactly, then the 
pressure on the surface of contact, ADB , will be of uniform 
intensity, and the case will be similar to that of fluid pressure 
on a cylinder. Let p denote this intensity whose direction is 
normal to ADB at every point of it (friction is omitted from 




3 2 4 


DETAILS OF CONSTRUCTION. 


consideration), then the total pressure in the direction of the 
arrow at K exerted by the plate on the rivet for each unit of 
length of the latter is equal to pAB. 

Let t be the thickness of the plate, as shown, and put d for 
the diameter AB, then the total pressure against the rivet in 
the direction of the arrow is 

P — ptd. 

The quantity p is the greatest mean value of the intensity 
of compressive stress which it is desirable to put upon the 
material under the given circumstances. It is usually taken 
as high as 12,000 lbs. per square inch, although 10,000 is 
a safer value. Of course, the actual maximum value of the 
intensity immediately in front of the centre, C, is much 
greater than either 10,000 lbs. or 12,000 lbs. If T' is the 
amount of stress which the joint is required to carry, then 
the number of rivets, so far as the previous consideration is 
concerned, is equal to 

T 

ptd ~ n ' 

The riveted joint itself, as shown in Fig. 5, PI. XI., may 
now be examined. The distance c should be at least 2]/ 2 
diameters of the rivet. By the arrangement of the rivets 
shown, when the pitch is from four to eight diameters, the 
strength of the plate of the width w will only be decreased 
by about the amount of metal taken out in one rivet-hole, 
although experiments to settle this point definitely are 
wanting. 

After having determined the pitch and distance of c , as 
above, there are five methods of rupture of the joint only 
which need serious attention. These five are: (1) tearing of 
the plate though the rivet-hole E, (2) tearing of the cover- 
plates through the rivet-holes at the middle of the joint, two 
in the figure, (3) shearing of the rivets, (4) and (5) rupture by 
compression at the surface of the contact between the rivets 
and the plates. The safe shearing stress to which rivets are 
subjected in bridge structures is usually taken at 7,500 lbs. 
This gives a safety factor of from 5 to 6. 



RIVETED„ CONNECTIONS. 


325 


Put 5 for the intensity of the maximum safe shearing stress 
on rivets (7,500 lbs. for wrought iron),/ for the intensity of 
the maximum compressive stress (10,000 to 12,000 for wrought 
iron), and T for the maximum working tensile stress; also, n! 
for the number of rivets on the line through the middle of the 
joint (two in the figure). Let t and t' represent the thickness 
of the plate and covers as shown. Then equal liability to 
rupture in the five ways mentioned is expressed as follows: 

7 T7ld 2 

Tt 0 w-d) = 2 Tt' (w-n'd) = 2. S = ntdp = 

4 

2 nt' dp — T'. 

It almost always happens that these quantities are not each 
equal to T\ but none of these should be less. If only one 
cover-plate is used, the 2 Tt ' should be replaced by Tt ', 2S by 
S, and 2 nt' by nt'. The form of this Equation of condition 
may be somewhat changed by piling of the plates, etc., but it 
will remain essentially the same, and serves to illustrate the 
principle which must govern in all cases. 

We see, therefore, that in obtaining the amount of pressure 
which should be put upon a rivet, / ought to be multiplied by 
its diameter, and not by the semi-circumference ADB, Fig. 1. 


Art. 71.—Riveted Connections between Web Members and Chords. 

When web members, as A and B , Fig. 6, PL XI., are riv¬ 
eted to the chord, the centre line {i. e ., the line joining the 
centres of gravity of the sections of the members) should pass 
through the centre of gravity of a system of points situated at 
the centres of the rivet-holes; otherwise the intensity of stress 
in any section of the member will not be uniform, and it (the 
web member) will be subjected to flexure. It is supposed, 
of course, that each rivet carries the same amount of stress, 
which, however, is probably seldom true, but it is the best 
assumption that can be made. Fig. 6 represents a proper dis¬ 
tribution of rivets in reference to the centre lines Ac and Be. 

It will be observed that the strut, composed of two un¬ 
equal legged angles, has its connection with the chord 
through both legs by means of the angle lugs. This should 



326 


DETAILS OF CONSTRUCTION. 


always be done in similar cases, for in no other way can an 
angle-brace develop its full strength. The practice of rivet¬ 
ing single legs, only, of angle-braces to chords is highly ob¬ 
jectionable, for the reason that the actual resistance of the 
brace is far below the nominal. 

When three or more pieces are riveted together at the 
same joint, all the centre lines of stress should intersect at 
one point, if flexure is to be avoided. It is frequently im¬ 
practicable to do this in riveted connections, and the imprac¬ 
ticability constitutes a serious objection to that character of 
work. 

If T is the total tension in the member A of Fig. 6, PI. XI., 
and C the compression in B , then there will be developed at 
a the bending moment : 

T x ac sin 6 ; 

and at b the bending moment: 

C x be sin 6 ; 

it being supposed that ac> be , and ab are the centre lines of 
stress of the two members and chord. 

It is very true that the metal is well supported in the 
vicinity of the joint, but unless provision is made for the flex¬ 
ure, as shown in Art. 69, which is seldom or never the case, 
some of the metal will be over-strained. Hence, this flexure 
should always be made a minimum, and reduced to zero if 
possible. 

In pin connections this bending at the joints is, of course, 
entirely obviated when all centre lines of stress intersect at 
the centre of the pin. 

Art. 72.—Floor-Beams and Stringers.—Plate Girders. 

The load applied to a bridge rests immediately on the floor- 
beams, generally speaking, and is transferred through them 
to the joints of the truss. In railway bridges the track and 
ties lie on stringers, which rest on the floor-beams. There are 
two or more stringers for each track. 

In highway bridges, the floor-beams support stringers, say 


FLOOR-BEAMS AND STRINGERS. 


327 


two feet apart (sometimes less and sometimes a little more), 
running parallel to the centre line of the bridge, which carry 
the floor. 

In railway bridges, the moving load is applied to the 
beams at the ends of the stringers, but in highway bridges 
the greatest moving load, for which the beam is to be de¬ 
signed, may be taken as uniformly distributed over the entire 
length of the beam, or rather that part of it between the 
points of support. 

Floor-beams should always be supported at the ends by a 
single hanger, or by some equivalent arrangement, which rests 
at the centre of the pin, or centre of the chord in riveted con¬ 
nections. Double hangers may be made tolerable by some 
equalizing device, usually of an expansive character, but as a 
rule they cannot be too strongly condemned, for in such cases 
the deflection of the beam will throw the greater part or all 
of the weight of the beam and its load on the inner hangers. 
The result will be not only a great overstraining of the latter, 
but a prejudicial redistribution of stresses in both web mem¬ 
bers and chords. The excessive load on the inner hangers 
will cause an overstrain in the inner tension braces which will 
extend to the inner lower chord members, and even to the 
posts and upper chord. 

Floor-beams are frequently built into vertical posts. In 
such cases an essentially central bearing on the pin should 
be provided. 

Plate girder stringers for railway bridges are either sup¬ 
ported in between the floor-beams, or partially so and par¬ 
tially above, or are supported wholly on the top of the floor- 
beams. With proper designing there is little difference in 
cost in the various methods. The first, however, is far prefer¬ 
able, for the reason that it gives the greatest stiffness to the 
floor system, other things being equal. 

The depths of plate girder stringers is usually found be¬ 
tween one-ninth and one-twelfth of their spans, 1. e the 
panel lengths. The floor-beam depth should be as great as 
economical considerations will permit, in order that the de¬ 
flections may be the smallest possible. 


3 2 8 


DETAILS OF CONSTRUCTION. 


The webs of stringers and floor-beams slightly aid resistance 
to flexure, but rivets in stiffeners and splice plates, if such ex¬ 
ist, decrease this resistance to some extent. Hence, it is the 
best practice to disregard the resistance of the web to flexure, 
and to assume that it resists the shear only. This is the more 
advisable when it is remembered that the rivets, in giving 
stress to the flanges, produce a flexure in the latter which is 
always neglected. This flexure arises from the fact that the 
rivet holes never pass through the centre of gravity of the 
flange angle section. 

The true depth of a plate girder is the vertical depth be¬ 
tween the rivet hole centres, but by a curious confusion 
between rolled and built sections, it is commonly taken as the 
vertical distance between the centres of gravity of the flange 
angles. 

All stress is given to the flanges by, or through, the rivets, 
binding them to the web, hence their proper distribution be¬ 
comes a matter of importance ; it will be shown by two ex¬ 
amples. 

The exact analytical determination of the web thickness 
cannot be reached, but the following approximate analysis is 
frequently used. 

It is shown by the theory of elasticity that if two planes 
at right angles to each other and to a plane normal to the 
neutral surface of a bent beam, be so taken that their intersec¬ 
tion shall be found in that neutral surface while their common 
inclinations to it LS45 0 , then there will exist at the neutral axis 
the same intensity of stress on the two planes, but one stress 
will be tension and the other compression. It is farther shown 
that the common intensity of the two stresses is the same as 
that of either the transverse or longitudinal shear at the same 
point, which is also known to be f the mean for the whole 
section in the case of a solid rectangular beam. 

Now, since the intensity of shear at the neutral surface of 
such a beam is a maximum and zero at the top and bottom 
surface, and since it has been assumed that the entire web 
takes the shear only, it follows that if the shear be assumed 
to be uniformly distributed throughout any transverse section 


FLOOR-BEAMS AND STRINGERS. 


329 


of the web, that the latter may be supposed to be composed 
of an indefinitely great number of columns, each of which is 
an indefinitely thin strip of the web, making an angle of 45 0 
with the axis of the beam. In a direction normal to these 
columns an equal intensity of tension will, of course, exist. 

One of the preceding assumptions is an error on the side 
of danger, by making the shear at the neutral surface only 
two-thirds of its actual value ; while the other, by making 
the shear at the top and bottom surfaces equal to the mean, 
instead of zero, is an error on the side of safety, and its in¬ 
fluence largely predominates over the former. 

The elementary columns of the web may be assumed to 
have their ends fixed at, or by, the flange rivets of a built 
beam, and if d' is the vertical depth, between rivet hole 

centres of the two flanges, the length of the elementary 
columns will be : 

l— d' sec 45 0 = 1.414 d' .(1). 


If 5 is the greatest total shear at any transverse section, 
A the area of that section of the web ; then taking the depth 
as d\ and s the mean shear, or: 




5 

A 5 


these elementary columns will be subjected to an intensity 
of compression equal to Hence if t, the thickness of the 
web, is sufficiently great, there may be taken by Gordon’s 
formula: 


s = 


/ 


1 + 


a P 




t= l \ 


f 


a(f-s) 



If, for wrought iron, a = 3000 and /= 8000, there will re- 
suit : 

t — 0.0183 l if ___ 

r 8000 


(4)- 












330 


DETAILS OF CONSTRUCTION. 


The empirical constants for steel are yet to be determined. 

In applying Eq. (4) to wrought iron plate girders, it will 
be found that the resulting values of t are excessive for large 
beams. The approximations already indicated, and the ad¬ 
ditional fact that the elementary columns are held in place 
throughout their whole length by the tension in the web, 
equal in intensity to the compression, and at right angles to 
the latter, are sufficient to justify the anticipation of such 
results. Eq. (3), therefore, has its chief value as the basis of 
an empirical formula for the web thickness. 

Although the web resists “ shear,” it is evident, from the 
preceding analysis, that the method of failure of a web will 
be that of buckling, in which the corrugations will be at right 
angles to the elementary columns. Hence, if the web is so 
held that these corrugations are prevented, its resistance will 
be very materially increased. This is accomplished by rivet¬ 
ing angles, usually in pairs, on each side of the web at proper 
intervals, so that the web plate is securely held between 
them. The office of these “stiffeners,” it is to be remem¬ 
bered, is simply to stiffen the web, and prevent its buckling. 
If they are assumed to act as struts, the transverse shearing 
strain in the web at the section considered, must in so much 
exceed the compressive strain in the stiffener-struts, that the 
rivets can transfer to it its proper load, at the same time pre¬ 
supposing a perfect condition of riveting. In reality, neither 
of those conditions can possibly exist. 

These stiffeners are ordinarily riveted to the web at right 
angles to the axis of the beam. If no elementary column is 
to be without the support of, at least, one of these stiffeners 
in some portion of its length, they must be placed at a dis¬ 
tance apart, measured along the axis of the beam not greater 
than the vertical depth between rivet hole centres ; and that 
limit is very commonly given in specifications, although it is 
sometimes placed at once and a half that depth. 

About the same amount of stiffening would be secured by 
placing the stiffeners at 45 0 with the axis of the beam, and 
at intervals of twice the depths, but the difficulties of con¬ 
struction would be increased. 


FLOOR-BEAMS AND STRINGERS. 331 

A safe rule, and one frequently used, though purely con¬ 
ventional, is to introduce stiffeners when the mean shear, i. e. y 
S -+-A, exceeds 4000 pounds per square inch for wrought iron. 

The function of the rivets holding the flanges to the web is 
next to be considered. In reality, these rivets may have two 
offices to perform. If the load of the girder rests on one of 
its flanges, the flange rivets will sustain it directly; but 
their chief office is to give the flanges their proper stress. If 
then the stresses be determined for any two points between 
the end of a girder, and the point of greatest flange stress, 
the shearing or bearing resistance of all the rivets between 
those points must‘be equal to, or not less than the resultant of 
the difference between the determined flange stresses and 
the load resting on the flange between the same points. If 
no load rests on the flange, but is carried directly by the web, 
the “ resultant ” is evidently the simple difference between 
the determined flange stresses. 

These elementary considerations constitute the entire 
method of finding the pitch and number of rivets in the 
flanges of built beams, and will be applied to two examples. 

The complete design of the truss, treated in Art. 11, is to 
be given in subsequent pages, and in the present connection 
the stringers and floor-beams will be discussed. The strin¬ 
gers will be placed 7 feet apart centres, and the rails will be 
laid on 8 inch by 8 inch ties 9 feet long, spaced 16 inches 
from centre to centre. The ties, rails, guard rails, splices, 
spikes, etc., will then weigh about 325 pounds per lin. ft. 
The depth of stringers will be taken at 27 inches throughout 
their lengths, and the iron of each stringer will be assumed 
to weigh 100 pounds per lineal foot. The total fixed load 
will then amount to 263 pounds per lineal foot of each strin¬ 
ger, while the moving load is one-half of the concentrations 
given in the engine diagram of Art. 11. The principles es¬ 
tablished in Art. 7 show that the four 10,000 pound driving 
wheel loads will produce the greatest bending moment when 
either v/heel A or A is at the distance 1.062 feet from the 
centre of the panel, and will be found under the wheel in 
question. 


33 2 


DETAILS OF CONSTRUCTION. 


With that position, the following moving load bending 
moments will exist: 


Maximum.1 19,200 ft. lbs. 

feet from the end.105,400 “ 

5 “ “ “ “.85,600 “ “ 

21 “ “ “ “.44,800 “ “ 


It is evident that the last three of these values are not the 
greatest moments for those points, but as the flanges are to be 
of uniform section, it is not necessary to seek them, as might 
easily be done by the aid of the principles of Art. 7. 

The vertical depth between the rivet hole centres of these 
stringers will be taken at 2 feet. The flange stresses at the 
various points will then be: 


At centre 


x (20.5 O 2 




7 i ft. from end ** X ™ * 1^5 + 105^00 = ^ . ER 


2X2 


<< u u 


26 3 -.x. 5 x I 5 J 5 + ^6oo = 47j920 « . GH ' 


2X2 


( 263 X 2 . 5 X 18.05 J 44joo = 6 tt ' KJL 

2 2X2 2 


The allowed working stresses in the flanges of the stringers 
will be taken at 7,000 pounds per sq. in. of gross section in 
compression and 8,000 pounds per sq. in. of net section in 
tension. The diameter of rivets in the stringers and floor 
beams of railway bridges is chiefly a matter of judgment ; it 
usually ranges three-quarters to seven-eighths of an inch. In 
the present instance, rivets of the latter diameter, before 
being driven, will be taken. The metal punched out for a 
rivet should leave a hole not more than one-sixteenth of an 
inch greater in diameter than that of the cold rivet. But 
the metal immediately about the edge of the hole is materi- 













FLOOR-BEAMS AND STRINGERS. 


333 


ally injured for tensile purposes, and in the tension-chord 
angles the disc of metal rendered valueless should be taken 
one-eighth of an inch greater in diameter than that of the 
cold rivet, i.e., for the present case, one inch. In the com¬ 
pression flange no metal need be deducted for the rivet 
holes. 

The upper flange section at the centre will be: 

66,540 A- 7,000 =9.5 sq. in. 

The net lower flange section will be : 

66,540 -f- 8,000 = 8.3 sq. in. 


The following flanges will satisfy the conditions : 


Upper flange 
Lower flange 


j 2 — 5" x 4" 30 lb. angles. 
( 1 — 10" x f" cover plate. 

2 — 5" x 4" 47 lb. angles. 


The 62 lb. angles have a thickness of three-quarters of an 
inch, so that the excess of gross section exactly covers the 
metal destroyed by the punch. 

It is the best practice to run a plate the whole length of 
the upper flange in order to make it act as a unit in resisting 
compression. The lower flange needs no cover plate, and, 
again, the additional rows of rivet holes would produce more 
dead metal. 

Although there is a little waste of metal near the ends in 
small plate girders, it is economy to save labor by making the 
flanges of uniform section throughout their lengths. This 
economy ceases when the flanges become so heavy that one 
or more cover plates become necessary in the tension flange 
and more than one in the compression flange. 

Cover plates should be carried at least a foot beyond the 
section at which the additional metal is required, and rivets 
should be closely pitched in that portion. If rivets pierce 
both legs of an angle in tension, metal should be deducted 
for all the rows in both legs. 


334 


DETAILS OF CONSTRUCTION. 


The location of the preceding sections is shown in Fig. I, 
of PL XI. The increments of stresses for the different seg¬ 
ments of the flanges will be: 

EC or FD — 66,540 — 59,144 = 7,396 lbs. 

GE or HF = 59,144 — 47,920 = 11,224 “ 

GK or HL - 47 , 9 2 ° — 25,367 = 22,553 “ 

KA or BL — = 25,367 “ 

The extent of flange over which the weight of each driving 
wheel weighing 10,000 pounds will be distributed is indeter¬ 
minate, but as the ties are 16 inches apart centres, it will be 
sufficiently accurate to take that distance as 2 \ feet. As 
each driving wheel passes over the entire stringer, the result¬ 
ant stresses which the rivets in the various sections into which 
the beam is divided will be obliged to carry, are as follows: 

EC or FD = y^( 7,396 f + (io,ooo) 2 = 12,440 

GE or HF = ^/(ii, 224) 2 + (io,ooo) 2 = 15,000. 

GK or HL — ^(22,553 f -t- (10,000) 2 = 24,700. 

KA or BL — ^(25,367^ + (io,ooo) 2 = 27,300. 

In order to determine the number of rivets in any of these 
sections, it will be necessary to fix the thickness of web plate. 
The minimum thickness permissible is, to some extent, a 
matter of judgment, but it is safe to say that no built beam 
for railroad purposes should have a less thickness of web 
than T 5 g of an inch, and a limit of is still better practice. 
The latter limit will be used here. 

With the position of moving load already determined in 
Art. II, the greatest shear at the end of the stringer is 
24,830 4- 2,700 = 27,530 pounds = X. The sectional area 
A of a 27 by f inch plate is 10.125 sq. ins. = A. Hence s = 
S ~ A = 2,720 pounds ; also 1 = 2 4 x 1.414 = 33.94 inches. 
Hence Eq. (4) gives: 

t = 0.45 inch. 


1 


Since this result is greater than the assumed value of t, the 






FLOOR-BEAMS AND STRINGERS. 


335 


hypothetical elementary columns are not capable of sustain¬ 
ing their loads without exceeding by a little the proper work¬ 
ing stress; but as the hypothesis involves a considerable 
safety error, the assumed value of t is probably ample. How¬ 
ever, as the additional metal is very small in amount, a pair 
of 3 by 2 \ 16-pound L stiffeners will be introduced at the 
distance of 27 inches from each end as shown. 

The working resistance to shearing offered by rivets in the 
truss will be taken at 7,500 pounds per sq. in., and the limit¬ 
ing pressure between rivets and walls of holes will be fixed at 
12,000 pounds per sq. in. under the same circumstances. The 
floor of a bridge is subject to shocks due to track imperfec¬ 
tions, and the above values should be reduced by 25 per cent., 
making the working resistance to shearing 5,625 pounds, and to 
pressure 9,000 pounds per sq. in. As the web is embraced by an 
angle iron on each side, each rivet will be subjected to double 
shear, and the thickness of bearing surface of the web will 
be much less than that of the two flange angles. In the de¬ 
termination of the shearing and bearing resistances of rivets, 
the cold diameter, before being driven, should, as a margin 
of safety, be considered. Hence, those resistances for the 
seven-eighths rivets under treatment, are: 


2.J. 7r (0.875) 2 • 5,625 = 6,750 pounds; and, 

0.875 . 0.375 . 9,000 = 3,000 “ 

The latter quantity is much the smaller, and will govern 
the number of rivets in each section, as follows: 


EC or FD .12,440 - 4 - 3,000 '= 

GE or HE. _15,000 4- 3,000 = 

GK or IiL .24,700 4 - 3,000 = 

KA or BL .27,300 -4 3,000 = 


4 

5 
8 

9 


rivets required. 

it u 

a u 

a U 


The nearest whole number is taken in each case. 

These results give at and near the end about nine rivets to 
each two and a half feet. A uniform pitch of three inches, 
therefore, will be assumed throughout each flange. If the 




336 


DETAILS OF CONSTRUCTION\ 


ioad is uniform in intensity, it is well known that the varia¬ 
tion of flange stress is very little for a considerable distance 
either side of the centre. This example shows, however, that 
concentrated loads may require just as close centre riveting, 
i.e., just as small pitch, as at the ends. The number of sec¬ 
tions into which a beam must be divided is a matter of judg¬ 
ment in each particular case. 

The greatest shear at the end of the stringer has already 
been seen to be 27,530 pounds, hence the end stiffeners at A 
must transfer that amount to the floor-beam web. An inten¬ 
sity of 4,000 pounds per square inch of normal section is a safe 
and proper value for those members. The end stiffeners will 
then be assumed to be 2 — 4" x 4" 35 lb. angles, one being 
on each side of the web at each end of the beam and extend¬ 
ing the full depth between the legs of the flange angles of the 
latter. Fillers whose thickness is just a little less than that 
of the heaviest flange angle will be required under these end 
stiffeners. Light intermediate stiffeners may be bent to fit 
if the flange angles are not too heavy; otherwise fillers must 
again be used. 

The number of rivets required to transfer the greatest re¬ 
action at the stringer ends to the floor-beam (see Fig. 2, PI. 
XI.) is found by taking the reaction of the locomotive load 
(found in Art. 11 to be 34,600 pounds) and adding to it the 
fixed load, or, 20.55 x 2 &3 — 5,405 pounds, then dividing the 
result by the bearing capacity of seven-eighths rivet in the 
three-eighths web of the floor-beam, as that is less than the 
resistance of the same rivet in double shear. The number of 
rivets needed will then be (34,600 + 5,405) 3,000 = 13. 

Seven rivets will be placed in each 4x4 end angle, as 
shown in Fig. 2, of PI. XI., making 14 for the end of each 
stiffener. If the loads were very great, it would be neces¬ 
sary to reinforce the web plate of the floor-beam where 
it receives the ends of the stiffeners, by riveting a plate 
on each side. In the present instance, however, it is un¬ 
necessary. 

The bill of material, with the weights of one stringer, may 
now be written as follows : 


FLOOR-BEAMS AND STRINGERS. 


33 7 


I 

27 

X 

3 

8 

Plate ' 20.55 ft. long... . 

....695 

lbs. 

2 

5 

X 

4 

30 lb. angles “ “ 

a 

. . . .411 

a 

2 

5 

X 

4 47 “ “ “ “ 

u 

. . . .644 

a 

I 

10 

X 

3 

8 

Plate “ “ 

a 

....257 

a 

4 

4 

X 

1 1 
Tlf 

Filling plates 19 ins. 

u 

.... 60 

a 

4 

3 

X 

1 

2 

“ “ 19 “ 

u 

... 32 

a 

4 

4 

X 

4 

35 lb. angles 27 “ 

a 

....105 

a 

4 

3 

X 

2i- 

^2 

16 “ “ 27 “ 

a 

.... 48 

a 




7 

8 

Rivets. 



a 


Total weight of stringer .2,372 lbs. 


The actual weight per foot is thus about 15 lbs. more than 
was assumed. Although this makes an insignificant differ¬ 
ence in the flange section, and is amply provided for in the 
present instance, in practice the actual weight should be a 
little under the assumed, and not over it. 

The arrangement of connecting the stringers to the floor- 
beams shown in Fig. 2 of PI. XI., has the merit of making a 
very stiff floor system. It is proper to say, however, that 
many other methods are used. The small angle brackets 
seen under the ends of the stringers and riveted to the lower 
flange angles of the floor-beam are simply for convenience in 
erection, and are not considered as essential to the resistance 
of the joint. 

According to the preceding bill of material, the actual 
maximum weight concentrated at the stringer ends will 
be: 

34,600 + 20.55 x 3°° — 40,800 nearly. 

The depth of the floor-beam will be taken at 36 inches 
throughout its entire length, so that the vertical distance be¬ 
tween rivet centres in the two flanges will be about 33 inches. 

If the weight of the floor-beam be assumed at 150 lbs. per 
lineal foot, the total flange stresses at the centre, at the 
stringer points, and at points 2 ]/^ feet distant from the ends, 
will be: 


22 













338 


DETAILS OF CONSTRUCTION . 


40,800 X 5 150 X (\ff _ 

2.75 8 x 2.75 


76,170, 


40,800 x_5 + 150x 5, XI2 = 75,840, 

2.75 2 x 2.75 


40,800 X 2.5 150 X 2. _ 5 x I4 . 5 = 38 0 9 0. 

2.75 2 X 2.75 


Using the same working stresses in the flanges as were 
fixed for the stringers, there will be found for the upper 
flange section at the centre: 

76,170 - 4 - 7,000 = 10.9 sq. ins.; 


and for the net lower flange section: 

76,170 - 4 - 8,000 — 9.5 sq. ins. 


The following flanges will satisfy these requirements: 


Upper flange 
Lower flange 


2 — 5" x 4" 36 lb. angles, 

1 — 12 x y 5 -g- cover plate. 

2 — 5" x 4" 55 lb. angles. 


The thickness of the 55 lb. angle is about 0.7 inch, so that 
if seven-eighth cold rivets are used the metal destroyed by 
the punch is just about equal to the excess of the total section 
over the net. 

The total shear at the end of the floor-beam is 40,800 + 8.5 
x 150 = 42,075 lbs. = 5 . If the web thickness be assumed 
at three-eighths of an inch, A — 36 x f = 13.5 sq. ins. 
Hence, s — S - 4 - A — 42,075 - 4 - 13.5 = 3,120 lbs. 

/ — 33 x 1,414 — 46.66. These quantities inserted in Eq. 

(4) give: 

t = 0.68 inch. 


This value shows that if the hypothetical elementary 
columns are to sustain 3,120 lbs to the sq. in., the web must 










FLOOR-BEAMS AND STRINGERS. 339 

be 0.68 inch thick. But such a thickness is plainly excessive, 
and shows how the formula errs in the direction of safety. 
A web thickness of three-eighths of an inch will be assumed, 
and 3" x 2J" 16 lb. angle stiffeners placed half-way between 
the stringer supports and the ends, as shown at EF, Fig. 2, 
PI. XI. These stiffeners will require 3" x •£" filling plates 28 
inches long. 

If shearing and bearing resistances of 5,625 and 9,000 lbs. 
per sq. in., respectively, be taken, as was done in designing 
the stringers, the bearing value of one rivet in the three- 
eighths web will be, as before, 0.875 x 0.375 x 9,000 = 3,000 
pounds. 

Hence the number of seven-eighths rivets required between 
the three sections of the beam will be: 


Centre and C .(76,170 — 75,840) -r- 3,000 = 1, 

C and EF .(75,840 — 38,090) -4- 3,000 =13, 

EF u AB .38,090 -r- 3,000 = 13. 


These' conditions will be sufficiently near fulfilled if a pitch 
of 2j inches be taken for a distance of 6 feet from each end, 
and six inches for the remaining 5 feet between the stringers. 
The flange stresses do not require a six-inch pitch at the 
centre, but that value should not be exceeded, in order that 
the flanges may be properly bonded. The pitch in the cover 
plate on the upper flange should be 2j inches for a distance 
of 15 inches from each end, and over the remaining portion 
of its length that pitch may be doubled. In no case, how¬ 
ever, should the pitch in a compression plate exceed about 16 
times its thickness. 

The end floor-beams are suspended from the pins by the 
plates shown riveted to the heavy end stiffeners, as at AB. 
Those stiffeners transfer half the weight of the beam in addi¬ 
tion to the 40,800 pounds at the stringer ends, to the suspen¬ 
sion plates, or, 40,800 + 150 x 8.5 = 42,100 lbs. As these 
end stiffeners do not in this instance sustain this entire weight 
at any section, the latter cannot be analytically determined. 
They should be heavy, however, and will be taken as 5 x 4 
36 lb. angles. 





340 


DETAILS OF CONSTRUCTION. 


The number of rivets required between the end stiffeners 
and the web is 42,100 -f- 3,000 = 14, but it is convenient to 
take 15 as shown. 

The dimensions of the suspension plates cannot be fixed 
until the diameter of the pin is determined, and they will be 
found in a later Art. 

The bill of material for the floor-beam with its weight will 
now be : 


1 36 x | Plate 17 ft. long 

2 5 x 4 36 lb. angles 

1 12 x T V cover plate 

2 5x455 lb. angles 
3 x | filling plates 2\ 


4 

4 

4 

4 

4 


u 


a 


u 


a 


u 


u 


u 


u 


n 


u 


u 


1 7 x 45 
34 x 12 
1 7 x 12.5 
34 x i8| 

9 \ x 64 


3 x 2J 16 lb. stiffening angles 3 ft. long 4 x 16 
5 x | end “ plates 2-|“ “ 10.4 x 8J 

5 x 4 36 lb. angles 3 “ “ 4 x 36 

3 x 2 \ 16 “■ “ 1 “ “ ij x 16 


rivets 


765 

408 

213 

623 

58 

64 

87 

144 

21 

130 


Total weight of beam .... 2,513 lbs. 

The weight per lineal foot is then 2,513 -f- 17 = 148 
pounds, or less than that assumed, as it should be. 

In the cases of large plate girders it is necessary to make 
splices in the web plate, a splice plate being used on each 
side of the web. The combined thickness of these splice 
plates should be at least 50 per cent, in excess of that of the * 
plates spliced. The number of rivets on each side of the joint 
should be such that the total bearing resistance in the web, or 
the total shearing resistance of the rivets, shall at least equal 
the greatest possible transverse shear at the joint considered. 
At least two rows of rivets should be found on either side of 
the joint. 

It is sometimes customary, if web plates have no splices, to 
take one-sixth of the web section as acting in either flange. 
If no rivet holes were punched for the stiffeners, this method 
would be allowable. But such rivet holes frequently take out 




FLOOR-BEAMS AND STRINGERS .. 


341 


considerable metal, and as the tension side of the plate only 
is affected, one-sixth of the remaining metal ceases to be a 
proper proportion. On the whole, therefore, it is better to 
neglect the bending resistance of the web, and allow it to 
balance, so far as it may, the effect of the rivet holes being 
out of the centre of gravity of the flange angles. 

That depth of plate girder which will give the least weight 
depends entirely upon the manner and amount of the load¬ 
ing. With very heavy concentrated loads, it may be half 
the span; on the other hand, with very light loads it may be 
less than one-twentieth of the span. As all bending moments 
may be supposed to be caused by some uniform load, either 
fictitious or real, the following analytical discussion may be 
of some value as well as interest: 

Economic Depth of Plate Girders with Uniform Flanges. 

If a plate girder carries a uniform load, and is designed 
with flanges of uniform cross sectional area, the depth which 
will give the least weight of girder may easily be obtained. 

Let / = span in feet. 

“ d — depth “ “ 

“ t — web thickness in inches. 

“ p — allowed working stress in lbs. per sq. in. for the flanges. 
“ p' — “ “ “ “ “ end stiffeners. 

“ a — sectional area of each intermediate stiffener in sq. ft. 
“ n — number of stiffeners (intermediate). 

“ w — total load per lin. ft. of girder (in pounds). 


The flange stress at centre will then be: 



The volume of the web plate in cu. ft. will be: 

Idf 

12 



342 


DETAILS OF CONSTRUCTION. 


If one-sixth of the latter is taken to be concentrated in the 
flange, the volume of the two flanges in cu. ft. will be: 


2 FI i Idt wP Idt 

_. _ 2_ — _ _ _ 

144/ 6 12 5 ?£>pd 36* 

The volume of the end stiffeners will be: 


wld 
144 /*’ 


and that of the intermediate stiffeners: 

nad. 

The volume of the entire girder will then take the value: 

V ~ —f—, + —5 + -—-} + nad .(5)o 

576/^ 18 144 p x 7 

By taking the first derivative; 
dV 


d{d) 


Solving for d 2 


d 2 = 


wl z It wl 

—-p — 73 + -x H- 7 + na = o. 

5 y6pd 2 18 144/ 


wP 


S- It wl \ 

576p ( -q + - t ? + na ) 

\i8 144 p x J 


.d = 


l 


wl 


2 \/ wl 


. ( 6 ). 


p{p>lt + ■—^ + 144^^) 


If one-sixth of the web is not concentrated in each flange, 
12# will take the place of 8 It in Eq. (6). 

If all stiffeners, both end and intermediate, are omitted, 
Eq. (6) will take form : 



In reality p is seldom or never exactly the same for both 




















EYE-BARS OR LINKS. 


343 


flanges, since it is the working stress in reference to the gross 
section. It will be sufficiently near, however, for all usual 
purposes to make it a mean of the two actual working stresses 
for the gross sections. 

It should be borne in mind that local circumstances fre¬ 
quently compel a different depth from that given by Eq. (6). 
It will also be found that a considerable variation from that 
depth will cause a comparatively small variation in weight. 
Again, the difficulties of handling a deep girder, and the shop 
cost per pound , may, and usually does, make the economic 
depth a little less than that found by the aid of Eq. (6). 


Art. 73.—Eye-Bars or Links. 

An “eye-bar” or “link” is a tension member of a pin- 
connection bridge, fitted at each end with an eye for the in¬ 
sertion of a pin. Two views of an eye-bar are shown in Fig. 
5, PI. XII.; A is the body of the bar, D the neck, and C the 
eye. The head of an eye-bar is the enlarged portion in which 
the pin-hole is made. The eye-bar is one of the most im¬ 
portant members of a pin-connection bridge, and the deter¬ 
mination of the relative dimensions of the head has been the 
subject of much experimenting. A mathematical investiga¬ 
tion, however, with the same object in view is a matter of 
considerable complexity, although an approximate solution 
of the problem may be obtained, and its agreement with the 
results of experiment is quite close. 

Before taking a general view of the stresses which may 
arise in an eye-bar head, it must be premised that a difference 
of -gV" to -gV' between the diameter of the pin and that of 
the pin-hole is considered exceedingly good practice. Before 
the eye-bar is strained, therefore, there is a line of contact 
only between the pin and eye-bar head, but on account of 
the elasticity of the material, this line changes to a surface 
when the bar is under stress, and increases with the degree of 
stress to which the bar is subjected. This line and surface of 
contact is, of course, in the vicinity of K , Fig. 3, PL XII., i.e ., 
on that side of the pin toward the nearest end of the bar. 
The consequence of this is that, when the bar is strained, the 


344 


DETAILS OF CONSTRUCTION. 


portion about KA, Fig. 3> is subjected to direct compression 
and extension; that about BL , DH, and FM to direct ten¬ 
sion and bending, while in the vicinity of CN (also CQ) there 
is a point of contra-flexure, and the stress in the direction of 
the circumference changes from compression to tension as H 
is approached from K. 

It should have been said before that if w represents the 
width of an eye-bar, as shown, then its thickness, t , is gener¬ 
ally included between the limits \w and These limits 

of the relative values of the quantities are seldom exceeded. 

Fig. 2, PL XII., represents a method of laying down an 
eye-bar head which has been determined by a very extensive 
system of experiments given by a member of the British In¬ 
stitution of Civil Engineers, and one that has stood the test 
of long American experience ; in short, there is probably no 
better method known. Let r represent the radius of the pin¬ 
hole, and w the width of the bar. 

Then take EN = o. 66 w. The curve DRBK is a semicircle 
with a radius equal to r + o. 66 w, with a centre, A, so taken 
on the centre line of the bar that QB — o&jw. GF is a por¬ 
tion of the same curve, with A' as the centre (A'C = A C ); 
67 /is any curve with a long radius joining GF gradually with 
the body of the bar. HG should be very gradual in order 
that there may be a large amount of metal in the vicinity of 
CG , for there the metal is subjected to flexure as well as 
direct tension. FD is a straight line parallel to the centre 
line of the bar. 

Fig. 3 shows another method founded on the results of a 
mathematical investigation. Take r and w as before. Then 
BC — AC — r + o .Syw, DH — §w — o. 66 w, FD = EF — 2r + 
w. DF is described with ED until DCF— 45 0 . BAB is de¬ 
scribed with BC until BCA = 35°. BN is drawn from A as a 
centre located in such a position as to cause that arc to be at 
the same time tangent to /Wand AB. DN is a straight line 
drawn parallel to the axis of the bar. PF is any easy curve 
which will appear the best. The dotted lines in both Fig. 2 
and Fig. 3 show the slope that should be given in order to 
clear a die. 


SIZE OF PINS. 


345 


The outline of the head is now usually formed of a portion 
of the circumference of a circle whose centre is the centre of 
the pin-hole. In such a case no dimension of the head should 
be less than the corresponding one determined by either of 
the methods just given. 

Fig. 4 shows the head thickened in such a manner that the 
mean maximum intensity of pressure between pin and pin¬ 
hole shall not exceed a given amount, p. Let T represent 
the maximum intensity of tension in the body of the bar; 
then, as has been shown in discussing the pressure against 
the bodies of rivets : 

wtT — 2 rpt .*. t — -. 

2 rp 

Art. 74.—Size of Pins. 

The exact analytical determination of the pin diameter in 
any particular case is, like many other matters, involving the 
elasticity of materials, an impossibility, although the problem 
in its simplest form was subject to a very able mathematical 
investigation by Charles Bender, C. E., in Van Nostrand's 
Magazine for October, 1873. One or two reasonable assump¬ 
tions, which, in a great majority of cases, must be very nearly 
accurate, give the problem a very simple character. The first 
of these assumptions is that the pressure applied to any pin has 
its centre at the centre of the surface of contact. Fig 3 of PI. 

XI. represents the half of a pin-connected joint, LL being the 

■ 

centre line, and by this assumption the centre of pressure be¬ 
tween each of the eye-bars A, B, C , D , etc., and post bearing 
P, and the pin is located half-way between the faces of those 
members normal to the axis of the pin. 

If, however, a pin is held by a compression member, such 
as an upper chord or post, then the centre of pressure in that 
member may be taken as the centre of such a surface as will 
reduce the bearing intensity to its maximum limit. 

It is to be premised that the general considerations touch¬ 
ing the distribution of pressure between rivets and plates 
given in Art. 70 hold equally true for pins. The greatest al- 



346 


DETAILS OF CONSTRUCTION . 


lowable bearing intensity between pins and eye-bars of 
wrought iron ranges from 10,000 to 12,500 pounds per square 
inch of the surface found by multiplying the diameter of pin 
by thickness of bar. The latter product is always considered 
the bearing surface. 

If two bars only, such as A and D, act on each end of a 
pin it is clear that the centre line of the latter will be convex 
toward D. The result will be a movement of the centres of 
pressure of those bars toward each other; so that the lower 
arm of A will be less than half the thickness of that member 
plus half that of B. The second assumption given above 
seems thus very reasonable, and may be extended to the case 
of a pair of eye-bars, only, at the end of a pin. When, on 
the other hand, a number of eye-bars of various sizes take 
hold of a pin, particularly if the bending moments have dif¬ 
ferent directions at different sections of the pin, the axis of 
the latter may be essentially straight and the centres of press¬ 
ure should be taken according to the first assumption. This 
is, in reality, the best practice in all cases, for if the centre of 
pressure departs from the axis of the bar, the latter will be 
subjected to a bending moment equal to the tension in the 
bar multiplied by the distance of the centre of pressure from 
its axis. Hence the necessity of so fixing the diameter of 
pin that it shall be as stiff as possible. 

In Fig. 3 of PI. XI., let a be the distance between the 
centres of eye-bars A and D; a', that between D and B ; a n 
that between B and E, etc., etc. These distances a, a!, a!\ 
etc., should always be taken as the thickness of the head plus 
one-eighth of an inch; the latter amount representing about 
the proper clearance in the best work. 

Then let T a , T d , T b , etc., represent the total tensions in the 
bars A , Z>, B, etc. The bending moments about the centres 
of those bars will then be: 

Aboiit centre of D ... .a T a 

“ “ “ B ...(* + a') T a - T d a\ 

“ “ “ E....(a + a' + a") T a + T h a" - T d (a' + a"). 

Etc., etc., etc., etc., etc. 


SIZE OF PINS. 347 

The rod R is a counter and does not usually act when the 
pin receives its greatest bending. 

The preceding moments are all similarly formed and are 
about vertical axes until the centre of the post bearing, P, is 
reached. The tension T of the main tension brace T oro- 

A 

duces a moment about an axis normal to its own. Let it be 
supposed that the resultant moment of all the chord members 
A, B , C, D and E about the centre of P is right-handed look¬ 
ing vertically down, as shown by M' in Fig. i. 

Let M' represent that moment by any con¬ 
venient scale. The moment of Thy a v , or 7 'a v , 
will be right-handed looking upward ; and let 
M t represent that moment by the same scale 
as before. The latter line is drawn normal to 
the axis of the member T. The line M will 
now represent by the same scale the moment 
to which the pin is subjected at the centre of 
P, and its direction is that of the axis of the 
moment. 

The greatest pin bending in the lower chord will usually 
take place with the greatest chord stresses, but the upper 
chord pins will receive their greatest moments by the great¬ 
est web stresses. 

When a number of bars are coupled to the pin in such a 
joint as that shown in Fig. 3 of PI. XI., it is usually necessary 
to test a number of sections in order to find the greatest mo¬ 
ment; unless the bars are very nearly of the same size and 
placed alternately as shown when the greatest moment will 
be found at the centre of the pin. 

It is frequently advisable, however, to employ different 
sized bars in order to reduce the bending moments; a small 
bar being placed at the end of the pin. 

The same reduction of bending moments is brought about 
even more effectually by the arrangement of lower chord bars 
shown in Fig. 2. 

In that figure it will be observed that the lower chord eye - 
bars are so grouped on any one pin, that the stresses in them- 
for each half of the pin, form couples which have opposite 




343 


DETAILS OF CONSTRUCTION. 


signs and, thus, to a great extent, or wholly, neutralize each 
other. 

By varying the sizes or thicknesses of the bars and by re¬ 
sorting to the method of grouping shown in Fig. 2 (which 



d b 

Fig. 2 . 


represents a portion of an actual lower chord) the bending 
moments in lower chord pins may easily be reduced to any 
desired extent in any case whatever. 

It is evident that the resultant moment shown in Fig. i 
could be obtained by resolving the stress T into its vertical 
and horizontal components and combining their moments with 
those of the lower chord stresses, making the components 
of M vertical and horizontal instead of vertical and inclined. 

The bending of pins is very much increased by thickened 
eye-bar heads, since the thickening increases the lever arm of 
the tensile stress in the eye-bar. The thickened eye is a most 
excellent thing for the bar, but necessitates an increased 
diameter of pin. 

The preceding operations illustrate the general method of 
finding the bending moment to which a pin is subjected in all 
cases; the component moments are determined from the 
stresses in the individual truss members, and the resultant is 
then found by the moment triangle or polygon. The pin 
diameter is then readily found in the following manner. 

If M is the external bending moment, / the moment of in¬ 
ertia of the normal section of the pin about its diameter D 
and K the intensity of stress in the fibres most remote from 
D } then it is known from the theory of flexure, since 1 





































SIZE OF PINS . 


349 


If K is known, Eq. (i) gives D at once after M is found by 
the general method exemplified by Fig. i, or in any other 
manner. 

For wrought-iron pins in the trusses of railway bridges, K 
is usually taken at 15,000 pounds. This value in Eq. (1) gives 
for wrought-iron pins: 

D = 0.089 */M t .( 2 ). 

For steel pins, under similar conditions, K may be taken at 
20,000 pounds, for which : 

D — 0.081 l/jf .( 3 ). 

There are numerous tables showing the bending moments 
of pins of all usual diameters with given values of K , so that 
in practice the computations expressed in Eqs. (1), (2) and 
(3) are seldom necessary. The value of M is determined for 
any particular case, after which, by the simple inspection of 
a table, the proper diameter may be chosen. 

It is seen by Eq. (1) that the diameter of a pin varies di- • 
rectly as the cube root of M and inversely as the cube root 
of K. 

It may sometimes happen that M T ,, in Fig. 1, is so small 
that it may be neglected ; in which case M — M'. 

No pin should possess a diameter less than eight-tenths the 
width of the widest bar coupled to it. 

When bending and bearing are properly provided for, a safe 
shearing resistance will be amply secured. If the apparent 
moment in the pin is sufficient to cause failure by flexure, it 
does not, by any means, follow that failure will actually take 
place; for the distortion of the pin beyond the elastic limit 
will relieve the outside eye-bars of a larger portion (in some 
cases perhaps all) of the stress in them. This result will pro¬ 
duce a redistribution of stress in the eye-bars, by which some 
will be understrained and the others correspondingly over¬ 
strained. Thus, although the pin may not wholly fail, the 
safety of the joint will be sacrificed by the overstrained metal 
in the eye-bars. 






350 


DETAILS OF CONSTRUCTION. 


Art. 75.—Camber. 


Camber is the curve given to the chords of a bridge, caus¬ 
ing the centre to be higher than the ends, or rather it is the 
amount of rise of the centre above the ends. It is given to a 
truss so that the chords may not fall below a horizontal line 
when the load is applied. Fig. 8, PL XII., represents a truss 
with exaggerated camber. The actual amount varies from 
-^th to ygVoth of the span. 

Camber may be given to a truss either by lengthening the 
upper chord or shortening the lower one; the latter method 
is preferable because the upper chord is sometimes not hori¬ 
zontal, and different panel lengths would have to be shortened 
by different amounts. 

On account of the unavoidable play at the joints of all 
work, the shortening of the lower chord, or lengthening of 
the upper, must be increased by about of an inch per 
panel in order to secure the desired camber. 

The lower chord shortening is made uniformly throughout 
its length ; that is, each panel length is shortened by a con¬ 
stant quantity. The true chords will, therefore, become arcs 
of circles of very large radii, and vertical posts will become 
radial. 

By means of the equation of the circle, y 2 = 2 Rx — x*, R 
being the radius, the amount of shortening or lengthening of 
chord to produce a given camber may be determined if the 
play at the joints be omitted. In the equation above, y rep¬ 
resents the half span and ;tr the camber desired, hence the 
radius 



2 x 


This is the radius of the lower chord when cambered, 

Generally it will be near enough to put R = — 

2 ^ 


The angular length of the lower chord will be 


a — 2 sin 


->,-1 z 

R 


= 2 sin 


. , 2 XV 

' }} 1 

y 1 + x l ’ 





CAMBER. 


351 


and the length in feet: 

l — Ra . 


The length of the upper chord will then be: 

* 

l = (R + d) a. 

The difference in length of chords will be: 

D = l' — l = da. 

This is the amount by which the lower chord is to be 
shortened, or the upper lengthened, in order to produce the 
required camber, if no play or strains exist. 

Since x is very small compared with y: 


a = 2 sin 


,-i 2x y _ 4 *y 


f + x z y 2 + x 2 


= (nearly). 


If the span is l x : 


a = 


8 x 

T" ; 


and, D = 


8 dx 

X* 


If r is such a ratio that d =rl 1 : 

D = 8r^r. 


On account of the play at the joints, should be taken a 
little-larger than the camber desired. 

Frequently r is about one-eighth, and for such a value: 



or, neglecting the play at the joints, the difference in lengths 
of the chords should equal the camber. 

If the chords are to be horizontal under the greatest loads, 
while T and C represent the supposed uniform intensities 
of tension and compression in the lower and upper chords 
respectively, E and E' representing the coefficients of elas¬ 
ticity ; 








352 


DETAILS OF CONSTRUCTION . 


This formula can only be approximate, for the chords are 
never exactly uniformly stressed, and the coefficient of elas¬ 
ticity is probably never the same throughout either chord. 

Since d, the depth of truss, does not vary, these formulae 

t 

apply only to trusses of uniform depth. 

A “ through ” truss has been supposed, but the same for¬ 
mulae exactly apply to a deck bridge. 

It is to be borne in mind that one-half the horizontal dis¬ 
tance between the centres of end pins is to be taken for y in 
determining R. If this distance is assumed in designing the 
truss, then the panel length is to be found by dividing / or /' 
by the number of panels. 

If the panel length is first assumed, and the camber pro¬ 
duced by shortening or lengthening zV, then this horizontal 
distance is essentially equal to the assumed chord length 
diminished or increased by D — da. 

In order to hold the camber in a truss, the diagonals must 
be shortened, as shown in Fig. 9, PI. XII. The diagonal 
which was bd before cambering, becomes ed afterward, ad 
and be are supposed to be panels in the upper and lower 
chords respectively before putting in the camber; afterward 
be becomes ef, while ad remains the same; the lower chord is 
supposed to be shortened. Let x be. the amount of shorten¬ 
ing of each panel of the lower chord = 2 be — 2 fc\ d, the 
depth of the truss; and p the original panel length equal to 
ad. Then 


ed — A^/dd + ee 2 = 




If the camber is produced by lengthening the upper chord, 
then ef is the original panel length, and ad the new one, and 


d ‘ + + ?)• 

In a triangular truss the diagonal gc Fig. to, is changed to 



gf- Vd* + xf. 







ECONOMIC DEPTH OF TRUSSES. 353 

If the upper chord is lengthened, eg is the diagonal desired, 
and ff the original panel length p. Hence, 

eg = V d* + i {p + x)\ 

Each diagonal is to be shortened to the length 'ed. 

In a draw-bridge each arm, in giving the camber, can be 
considered one span, but the whole amount of shortening in 
the lower chord of one arm must also be taken out of the up¬ 
per chord at the centre. If this is not done, the ends will sink 
below their original positions. 

Art. 76.—Economic Depth of Trusses with Parallel Chords. 

The so-called economic depth of truss for a given span, is 
that depth which involves the least material or weight of 
roetal in the bridge. This depth depends upon the intensity 
of moving load for each truss, the length of panel, the great¬ 
est allowable stresses, etc., etc. Various mathematical inves¬ 
tigations have been made with a view to the determination 
of this depth of truss in terms of the length of span. But 
on account of the exceedingly intricate character of the 
problem, any feasible analysis must be based upon assump¬ 
tions which simplify the analytical operations, but render the 
results only approximately true. These investigations, how¬ 
ever, and the experience of American engineers, show that a 
depth varying from one-fifth to one-seventh the length of 
span will give the least weight of truss; the former for very 
heavy loads, as in two truss double track bridges, and the 
latter for light loads. 

When the span becomes very long, i. e., 400 to 500 feet, 
the depth of truss increases to an unusual height, and the 
cost of erection is correspondingly large. The depth is then 
frequently taken not larger than one-eighth the span, or even 
less. 

Again, local conditions, such as the necessarily uniform 
depth (for the sake of appearance) of adjacent spans of vary¬ 
ing length, sufficient depth of short spans for over-head brac¬ 
ing (very necessary for lateral stability), etc., in the majority 

23 



354 


DETAILS OF CONSTRUCTION. 


of cases exclude the use of the economic depth, even if it 
were exactly known. 

It is to be borne in mind, also, that the lightest truss is not 
necessarily the cheapest. That bridge is the most economi¬ 
cal which can be made ready for traffic for the least money. 

Facility in working up details, and the least possible amount 
of time in the shop, are very important elements, indeed, in 
every design. 

In fact, the lightest weight does not make the most econom¬ 
ical bridge, for the reason that the shop cost per pound is 
greater than with a somewhat increased weight of metal. 
When it is borne in mind that a considerable variation may 
be made from the depth of least weight, without affecting 
that weight to any considerable extent (as actual computa¬ 
tions show to be the case), it is easy to understand that the 
truly economic depth is materially less than that which gives 
precisely the least weight of material. 

Long panels are an economic feature of any bridge possess¬ 
ing a system of floor-beams and stringers, as well as condu¬ 
cive to other points of merit. The resulting concentration of 
metal not only leads to less weight and rate of cost in the 
shop, but enhances, also, the stiffness and stability of the in¬ 
dividual members. 

For economy in weight, long panels require a greater depth 
than shorter panels. 

This much may be said in regard to continuous trusses: 
On account of the existence of the points of contraflexure, 
they require considerably less depth than trusses that are 
not continuous, used on the same points of support. The 
depth of the latter, therefore, is a limit which should never 
be reached by the depth of the former. 


Art. 77. —Fixed and Moving Loads. 

Both fixed and moving loads depend upon the local cir¬ 
cumstances of each case, and the former very much upon the 
character of the design. A depth from one-fifth to one- 
seventh the span will give a very light fixed weight, but a 


FIXED AND MOVING LOADS. 


355 


depth of one-twelfth the span will involve a considerable in¬ 
crease of weight, while the moving load remains the same. 

The weight of a single-track railway floor, for the present 
(1885) existing moving loads, may be taken at about 400 
pounds per foot. 

The moving load, also, depends upon the length of span. 
If the span is very great, the probability of the whole bridge 
being covered with an excessively heavy moving load is very 
slight, if any exists at all. If the span is short, one or two 
locomotives may cover the whole bridge, thus causing the 
moving load, per foot, to be very great for the whole span. 

Thus it is seen that the moving load, per foot, may decrease 
as the span increases. 

The whole matter of moving loads for both highway and 
railway bridges is well illustrated by the following tables, 
taken from “A Bill to secure greater Safety for Public Travel 
over Bridges,” introduced in the Sixty-second General As¬ 
sembly of the State of Ohio, shortly after the Ashtabula 
disaster: 


For City and Suburban Highzvay Bridges. 


Span 

in feet. 

Moving load per sqtiare foot. 

O 

to 

30. 

. 110 pounds. 

30 

u 

50. 

. IOO 

a 

50 

u 

75 . 


u 

75 

u 

IOO. 

. 80 

u 

100 

u 

200. 

. 75 

u 

200 

u 

400. 

. 65 

a 


All other Highway Bridges. 

Span in feet. 

o to 30 

30 “ 50 

50 “ 75 

75 “ 100 

100 “ 200 

200 “ 400 


Moving load per square foot. 

... 100 pounds. 

,.. 90 “ 

... 80 “ 

...75 “ 

60 “ 

... 50 “ 














356 


DETAILS OF CONSTRUCTION. 


Railway Bridges. 

Moving load per lineal 

Span in feet. ' foot of each track. 

o to y\ .. 9,000 pounds. 


7 i 

<< 

10 . 

.7.500 

<< 

10 

a 

I2i . 

.6,700 

u 

12 i 

a 

15 . 

.6,000 

u 

15 

a 

20 . 

.5,000 

a 

20 

a 

30 . 

. 4 > 3 °° 

a 

30 

a 

40 . 

. 3700 

a 

40 

a 

50 . 

. 3 > 3 °° 

a 

50 

(( 

75 . 

. 3,200 

a 

75 

a 

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. 3,ioo 

a 

100 

a 

150 . 

.3,ooo 

u 

150 

a 

200 . 


a 

200 

a 

300 . 

.2,800 

a 

300 

a 

400 . 


a 

400 

a 

0 

0 

VO 


u 


Floor-beams and stringers are really bridges of short 
spans equal to their lengths, consequently they must be 
designed for the heavy loads belonging to those short 
spans. Fig. i shows the locomotive weight specified by 



Fig. i. 


Mr. Theodore Cooper, C.E., in his “ General Specifica¬ 
tions for Iron Bridges and Viaducts,” while Fig. 2 shows 
the heavy passenger locomotive used by Mr. Jas. M. Wil¬ 
son, C.E., in his standard specifications for the Pennsyl¬ 
vania R. R. 
























FIXED AND MOVING LOADS. 


35 7 


These represent the heaviest engines of their type now in 
use. There is, however, a heavy decapod engine shown by 
Fig. 3, beginning to make its appearance. 



i 


Fig. 2. 

The moving load usually specified consists of two of any of 
these types of locomotives followed by a uniform train of 
3,000 pounds per lineal foot. Occasionally the locomotive 
concentrations are followed by those of the train. 

Besides the preceding heavy moving loads, there are in- 



Fig. 3. 

numerable lighter ones depending upon local circumstances 
of traffic. 

The actual concentrations play a very important part in 
bridge computations. The old method of a uniform load, 
even with an engine excess, no longer fulfils the requirements 
of the best engineering practice, particularly in the treatment 
of short spans. 

This is well illustrated by the following table which shows 
the uniform load per lineal foot which will produce the same 


1 









353 


DETAILS OF CONSTRUCTION. 


chord stresses at the centre of the span as the actual concen¬ 
trations shown in Fig. I. 


Span in feet. 

Equiv. uniform load 
in lbs. per lin.ft. 

Span in feet. 

Equiv. uniform load 
in lbs. per lin. ft. 

55 

3,750 

25 

4,838 

50 

3,866 

20 

5 ,i 37 

45 

4,004 

15 

5,760 

40 

4,242 

12 

6,000 

35 

4-336 

IO 

5,766 

30 

4,572 

5 

9,600 


Above 55 feet the equivalent uniform load per lineal foot 
will slowly decrease until it reaches a value of about 3,200 
lbs. for 100 feet and over, i. e ., supposing the moving load to 
consist of a train of such locomotives. 


Art. 78.—Safety Factors and Working Stresses. 

Although the subjects of safety factors and working 
stresses properly belong to the domain of the resistance of 
materials, they may here be touched upon in a general 
manner. 

The fixed weight of a long span bridge is much greater, 
per foot, than that of a short span. Again, it has been seen 
in the preceding article that the moving load for a long span 
is much less than that for a short span. For both these rea¬ 
sons, the variations of stress in passing from a loaded to an 
unloaded condition are much greater in the material of a 
short span than that of a long one. Consequently, the mate¬ 
rial will be much more fatigued in a short span than in a 
long one. 

Although the subject of the fatigue of metals is yet in an 
unsettled state, it is clearly established that these conditions 
of stress in short spans demand a larger safety factor, or 
smaller working stress, than those in the long spans. 

Again, in any bridge or truss whatever, carrying a moving 
load, some parts are subject to a much greater variation of 


SAFETY FACTORS AND WORKING STRESSES. 359 

stress in the process of first being subject to, and then relieved 
of, loads than others. 

Counter-braces may not be, and probably are not, strained 
at all by the fixed load ; but they take a proper working stress 
under the action of the moving load. 

The condition of loading for greatest stress in any main 
web member, except those at the ends, is a partial covering 
of the span. But the fixed load is distributed over the whole 
span. Hence the variation of stress in the main web mem¬ 
bers will be greatest at the middle of the span, and least at 
the end. At the centre, however, the variation is much less 
than in the counter-braces. 

The fatigue of the material, therefore, requires that the 
greatest safety factors , or least working stresses , be found in the 
counter-braces; and that the working stresses in the main web 
members at the centre be greater than those in the counter-braces, 
but less than those in the main web members at the ends of the 
truss . 

The disposition of the moving load for the greatest chord 
stresses is, in all cases, essentially the same as that of the fixed 
load. Hence the variation of stress will be essentially the 
same throughout the chords, and the safety factor or working 
stress may be uniform throughout each chord ; the safety 
factor being the same as that in the end web members sus¬ 
taining the same kind of stress. 

If a structure is to carry a fixed load only, the safety factor 
may be three for wrought-iron and steel, and possibly as small 
for good qualities of cast-iron and timber. As a rule, how¬ 
ever, cast-iron and timber require a larger safety factor than 
wrought-iron and steel. Local circumstances affect, to a 
great extent, working stress. If the risk (respecting life and 
property) attending failure is small, the safety factor may be 
small also. But if the risk is great, the safety factor must be 
correspondingly great. 

In the truss members of long span bridges of wrought-iron 
and steel, the safety factors may vary from three and a half 
or four to five; but in short spans of the same material, they 
should vary from about five to six or eight. 


3 6o 


DETAILS OF CONSTRUCTION. 


Good cast-iron should be found with safety factors varying 
from six to ten, while those for timber may vary from eight 
to twelve. 

It is not to be supposed from these large safety factors that 
the determination of the stresses or the character of the vari¬ 
ous materials is so excessively uncertain. It is certainly true 
that there is some indetermination in these respects, but only 
a little in comparison with that connected with the mode of 
application of the moving load. 

With a perfect condition of track, a rapidly moving train is 
supposed by many to approximate very closely to a suddenly 
applied load, although it is quite certain that it does not. 
For this reason some engineers have doubled the moving 
loads, in making their calculations, and then fixed the values 
of the safety factors as if all loads were gradually applied. 

But no track is in perfect condition, and all rough places, 
or lack of continuity, such as rail joints more or less open, 
produce shocks which cause greater stress than any suddenly 
applied loads. The amounts of these last stresses are inde¬ 
terminate, for the extent of their causes can scarcely be 
determined. 

Again, Mr. J. W. Cloud, C. E., at the Philadelphia meeting 
of the American Institute of Mining Engineers, February, 
1881, pointed out the existence of certain unrecognized 
stresses ; such as those caused by the vertical component of 
the thrust of the connecting-rod of a locomotive, which alter¬ 
nates in direction twice in each revolution of the driving- 
wheels, thus producing a pulsating effect, as well as those 
which arise from the lack of balance of the driving-wheels in 
a vertical direction. 

All these causes produce stresses which it is impossible to 
measure, and the safety factor must cover all uncertainties. 

It is possible that a more highly perfected track and the 
production of more nearly uniform material in connection 
with an extended experience may justify the reduction of 
safety factors. 

The following “Table of Tubular and Truss Bridges for 
Single and Double Track Railways, constructed of Iron and 


SAFETY FAC TOTS AND WORKING STRESSES. 361 

Steel and having Spans exceeding 300 feet,” gives the work¬ 
ing stresses, loads, and other interesting data of some of the 
principal bridges of the world. It is taken (with the excep¬ 
tion of No. 18) from the “ Proceedings of the Institution of 
Civil Engineers” of Great Britain, Vol. LIV. The greater 
portion of it is there given in connection with a paper by 
Mr. T. C. Clarke, on “ Long Span Bridges.” 


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fcfc i 4 

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In No. 18, the 422 tons are for iron and steel. The writer is indebted to the kindness of Mr Morison himself for the information in regard to this bridge. 













































































GENERAL OBSERVATIONS. 


365 


Art. 79.—General Observations. 

All abutting surfaces in bridges, or similar structures, 
should be very carefully machine finished. 

Where pins bear against portions of the upper chord, as at 
c, d , and e of Fig. 5, PI. III., the amount of bearing surface 
should be determined as for rivets, and sufficient area given 
by riveting on thickening plates, if necessary. A thickening 
plate is shown at the joints of the same figure. The num¬ 
ber of rivets for the thickening plate is determined by the 
amount of pressure allowed on each one, as has already been 
shown. 

If a finished piece is to fit into a finished cavity, however 
well the work may be done, there must be at least -fa inch 
“ play.” 

One end of a truss bridge, unless the span is very short, 
usually rests upon “ expansion ” rollers, from two to four 
inches in diameter. An approximate formula for the resist- 
ance of such rollers is given in the Appendix. 


CHAPTER XI. 


WIND STRESSES AND BRACED PIERS. 

Art. 80.—Wind Pressure. 

In a paper presented to the American Society of CivL 
Engineers (Transactions, Vol. X.), Mr. C. Schaler Smith gives 
some very valuable information in regard to wind pressure. 
The highest observed pressure which has come within his 
knowledge is 93 pounds per square foot. This pressure 
derailed a locomotive at East St. Louis, Mo., in 1871. In his 
own specification he says :— 

“ The portal, vertical, and horizontal bracing shall be pro¬ 
portioned for a wind pressure of 30 pounds per square foot 
on the surface of a train averaging 10 square feet per lineal 
foot, and on twice the vertical surface of one truss.” 

The wind pressure on a train is a moving load, and should 
be so considered, while the wind pressure on the trusses is a 
fixed load. 

His experiments on the Rock Island draw-bridge showed 
that the wind pressure against the two trusses was over 1.8 
times that on the exposed surface of one. 

Again, quoting from his paper:— 

“ The Erie specifications are as follow: 

Fixed load, roadway chord, 150 lbs. per lineal foot. 

“ “ other “ 150 “ “ “ “ 

Moving “ roadway “ 300 “ “ “ “ 

Iron in tension at 15,000 pounds. 

“ “ compression, factor 4. 

“ The Pittsburg, Cincinnati and St. Louis Railway requires 

366 


WIND PRESSURE. 


367 

300 pounds per foot for the train, and 30 pounds per square 
foot on one truss only. 

“ For the bridge over the Missouri, at Glasgow, 50 pounds 
per square foot on one truss, and 300 pounds per lineal foot 
of train were used. 

“ For the Eads bridge, at St. Louis, 50 pounds per square 
foot on the structure alone was the specified pressure. 

“ For the Kentucky River bridge the wind pressure was 
assumed at 31^ pounds per square foot on spans, train, and 
piers, and factor 4 was used in proportioning the bracing. 

“ The Portage bridge, New York, was built to resist 30 pounds 
per square foot on structure and train, and 50 pounds per 
square foot on the structure alone. 

“ The 520 feet span over the Ohio, at Cincinnati, was de¬ 
signed to withstand 50 pounds per square foot on structure 
alone, or 30 pounds per square foot on train and structure 
combined. 

“A fully loaded passenger train, and the heaviest possible 
freight train, will leave the track at the respective pressures of 
3I4 and 56^ pounds per square foot.” 

Engineers frequently specify 30 pounds per square foot of 
trusses and train combined, or 50 pounds per square foot of 
trusses alone. 

300 pounds per linear foot of single track is also frequently 
used for moving wind pressure on train. 

The following refers to the single track bridge at Platts- 
mouth, Neb., and is from the Railroad Gazette , 17th Dec., 
1880: “The structure is also designed to resist a lateral wind 
pressure of 500 pounds per lineal foot on the floor, and 200 
pounds per lineal foot on the top chord of the through spans 
and the bottom chord of the deck spans; these quantities are 
about equivalent to a wind pressure of 30 pounds per square 
foot on the bridge when covered by a train, and to 50 pounds 
per square foot on the empty bridge.” 

The following are a set of rules recommended for English 
practice, almost exactly in the words of the report: 


368 


WIND STRESSES AND TRACED TIERS. 


Report of the Committee appointed to consider the Question 
of Wind Pressure on Railway Structures , by the Board of 
Trade of London, made on the 20th May, 1881. 

The following rules were recommended : 

(1) . For railway bridges and viaducts, a maximum pressure 
of 56 pounds per square foot should be assumed for purposes 
of calculation. 

(2) . That when the bridge or viaduct is formed of close 
girders, and the tops of such girders are as high or higher 
than the tops of passing trains, the total wind pressure upon 
such bridge or viaduct should be ascertained by applying the 
full pressure of 56 pounds per square foot to the entire verti¬ 
cal surface of one main girder only. But if the top of a 
train passing over the bridge is higher than the tops of the 
main girders, the total wind pressure upon such bridge or 
viaduct should be ascertained by applying the full pressure 
of 56 pounds per square foot to the entire vertical surface, 
from the bottom of the main girders to the top of the train 
passing over the bridge. 

(3) . That when the bridge is of the lattice form, or of 
open construction, the wind pressure upon the outward or 
windward girder should be ascertained by applying the full 
pressure of 56 pounds per square foot, as if the girder were 
a close one, from the level of rails to the top of the train 
passing the bridge or viaduct, and by applying, in addition, 
the full pressure of 56 pounds per square foot to the as¬ 
certained vertical area of surface of the iron work of the 
same girder, situated below the level of the rails or above 
the top of a train passing over such bridge or viaduct. The 
wind pressure upon the inward or leeward girder or girders 
should be ascertained by applying a pressure per square foot 
to the ascertained vertical area of the surface of the iron 
work of one girder only, situated below the level of the rails, 
or above the top of a train passing over the said bridge or 
viaduct, according to the following scale : 

(a). If the surface area of the open spaces does not 
exceed f of the whole area included within the outline of the 


WIND PRESSURE . ^69 

girder, the pressure should be taken at 28 pounds per square 
foot. 

($). If the surface area of the open spaces lie between | 
and |- of the whole area included within the outline of the 
girder, the pressure should be taken at 42 pounds per square 
foot. 

(c). If the surface area of the open spaces be greater than 
| of the whole area included within the outline of the girder, 
the pressure should be taken at 56 pounds per square foot. 

(4) . That the pressure upon arches and piers of bridges 
and viaducts should be ascertained, as nearly as possible, in 
conformity with the rules above stated. 

(5) . That in order to insure a proper margin of safety for 
bridges and viaducts, in respect of the strains caused by wind 
pressure, they should be made of sufficient strength to with¬ 
stand a strain of 4 times the amount due to the pressure 
calculated by the foregoing rules. And that for cases where 
the tendency of the wind to overthrow structures is counter¬ 
balanced by gravity alone, a safety factor of 2 will be 
sufficient. 

John Hawkshaw. 

W. G. Armstrong. 

W. H. Barlow. 

G. G. Stokes. 

W. Yolland. 

The evidence before us does not enable us to judge of the 
lateral extent of the extreme high pressures occasionally re¬ 
corded by anemometers, and we think it desirable that 
experiments should be made to determine this question. If 
the lateral extent of exceptionally heavy gusts should prove 
to be very small, it would become a question whether some 
relaxation might not be permitted in the requirements of 
this report. 

W. G. Armstrong. 

G. G. Stokes.” 

Up to the date of the above report, the highest pressure 
per square foot ever recorded at Glasgow was 47 pounds ; 

24 


370 


WIND STRESSES AND BRACED PIERS. 


while the highest ever recorded at Bidston, near Liverpool* 
was 90 pounds per square foot. At another time, at Bidston, 
80 pounds per square foot was recorded. 

The above committee also found that, if P is the maxi¬ 
mum pressure per square foot, V the maximum run of wind 
in miles per hour, both these quantities being observed 
by anemometers, the following equation very nearly held 
true : 



100 


Art. 81.—Sway Bracing. 

The construction of the upper and lower sway bracing of 
a truss must, so far as the jar and oscillation of a moving 
road are concerned, be a matter of judgment; but the stresses 
due to the action of the wind may be determined with suffi¬ 
cient accuracy. 

Although a moving train will partially shelter one truss, it 
seems no more than prudent, with the ordinary open style of 
American bridge, to consider the action of the wind as exist¬ 
ing constantly, during the passing of a train, over the whole 
of the projection of each truss in the bridge on a plane nor¬ 
mal to the direction of the wind. If this is considered exces¬ 
sive, however, for low trusses, that portion of the windward 
truss sheltered by the train may be omitted. 

Let Fig. 1 and Fig. 2 represent a single concellation rail¬ 
way truss bridge, with vertical and diagonal bracing, and 
let the wind be supposed to blow in the direction shown 
by the arrow, which is normal to the planes of the trusses. 
Primed letters belong to the truss DC'N'O ,' but all are not 
shown. 

As the truss is a “ through” one, all wind pressure against 
the floor system will act in the lower chord. 

With the wind pressure between thirty and forty pounds 
per square foot, the following loads may be taken at the 
various panel points : 


SWAY BRACING. 

At C, G\ K\ L\ M', N\ C, G, K, L, M, N. 

“ D and O' . 

“ Intermediate points. 

“ B and 0 . 

“ Intermediate points. 




The amount 3.01 tons involves the pressure against the 
train, which is taken at 300 pounds per foot of track. The 
panel length is fourteen feet, hence the panel train load is 
14 x 300 = 4,200 pounds = 2.10 tons. The wind pressure 
against the floor system is assumed to be 0.56 ton per panel, 
while the panel pressure against each truss is 0.35 ton. The 
sum of the three quantities is 3.01 tons. 

The panel train loads (2.10 tons) constitute a continuous, mov¬ 
ing load; the wind pressure against the trusses and floor sys¬ 
tem, however, forms a fixed load. 

The following are the truss dimensions, including the 
lengths of the braces AF and A'F' : 

Panel length = 14.00 feet. Height of truss = 16.00 feet. 
Width, BD = 14.00 “ BC = 21.26 “ 

CF — 7.00 “ AC = 4-°4 “ 


371 

0.35 tons. 
0.18 “ 
0.35 “ 

0.18 “ 


F A = 8.08 feet. 























372 


WIND STRESSES AND BRACED TIERS. 


Normal from C on FA — CF x sin 30° — 3.5 feet. 

Let H represent half the total wind pressure concentrated 
in the two upper chords; this will be resisted (if the bridge is 
not blown bodily off the abutments or piers) by an equal 
force of friction developed at the feet B and D, or 0 and O' 
of the end posts. Let H' and H" be the forces developed at 
D and B , lespectively. These horizontal forces will tend to 
overturn the trusses in a vertical plane normal to the axis of 
the bridge. A vertically upward reaction, V, will be devel¬ 
oped at B, and an equal downward one (a portion of the 
weight of the truss DC'N'O') at C'. Considering the left 
end of the truss, the following three conditional equations of 


equilibrium must be fulfilled : 

H'+H n +H= o ....... (1). 

V+V'=o ....... (2). 

+ 16+Vx 14 = 0 .(3). 


The vertical force acting at C is represented by V. 

These three equations are not sufficient for the determina¬ 
tion of the four quantities H\ H", V, and V'. The forces H' 
and H n are therefore indeterminate in magnitude, except in 
this respect, their sum must be equal and opposite to H. 

With the form of portal bracing shown in Fig. I, it will be 
assumed in this article that H" = o and H f — — H. Other 
and better forms of portal, together with other assumptions 
in regard to the horizontal reactions H' and H ", will be given 
in succeeding articles. 

From the data already given: 

H — 6 x 0.35 = 2.10 tons. 

Hence by Eqs. (3) and (2) ; 

V — 2.40 tons = — V. 

I 

At the point E’ let there be supposed to act two forces 





SWAY BRACING. 


373 


equal, opposite, and parallel to H and H'; these two forces 
will balance each other. Instead of the two forces H and 
H\ there may then be taken two couples, M' — H' x 14, and 
M= fix 16. 

In the same manner at E let two forces equal, opposite, 
and parallel to Fand V be supposed to act. Then, instead 
of V and F, there will exist cwo couples, M" = V x 14, and 
AT" = V x 14. 

The couples whose moments are M and M"' balance each 
other, as is shown by Eq. (3). The couples whose moments 
are M 'and M" have axes at right angles, consequently their 
resultant will be: 


M t = \/M ,2 + M" 2 = 14 V (2.i) 2 4 (2.4) 2 = 44.66 ft. tons. 

The plane in which M 1 acts, contains the chords BO and 
C'N’, and the direction of the couple is such that it causes 
compression in C'N’ and tension in BO. Consequently for 
those stresses 


(BO) — — (C'N') = 44.66 -f- 21.26 = + 2.1 tons. 

As a check ; (BO) =4 V x 14 4- 16 = + 2.1 tons. 

The following stresses in the members of the portal of the 
bridge may now be written : 

(A'E') — H x 21.26 -4- 3.5 = 4 12.8 tons. 

(A'C') = — 12.8 x sin 30° = — 6.4 “ 

(F'C r ) — — 12.8 x cos 30° = — II. 1 “ 

The greatest bending moment in DC’ exists at F', and is: 

M 2 = 2.1 x (21.26 — 7.00) = 29.95 ft. tons. 


The stress in BC is: 


(BC) = - 


21.26 


x H = 


21.26 

16 


x V = — 3.19 tons. 






374 


WltiD STRESSES AND BRACED PIERS. 


The compressive stress in DC\ due to the vertical loading, 
is relieved by the same amount. 

The greatest bending moment in CC' exists at A', and has 
for its value : 

M 3 = (BC) X (14.00 — 4.04) = 31.77 ft. tons. 


With the wind in the direction taken, the brace AF must 
be supposed not to act at all. Both moments and M 3 pro¬ 
duce bending in the plane of the portal. 

The end post (. DC ), always of uniform cross section, must 
be able to resist with a proper safety factor, at F', the bend¬ 
ing moment M 2 . The sway brace CC' must be able to resist 
the moment M 3 at both the points A and A'. 

The ordinary truss stresses in the sway bracing remain to 
be found. 

In both upper and lower sway bracing the inclined members 
are tension ones only, while those normal to the planes of the 
trusses (in the direction of the wind) sustain compression 
only. 

In the upper chord the truss C'GMN' has the two 
points of support C' and N'. The following trigonometric 
quantities will be required : 


Angle G'KK ' =45° tan 45 0 = 1. 

sec 45 0 = 1.414. 

The upper web stresses are the following: 


(K/C) = 





~ — 

0.35 

tons. 

(G'K) = 

+ 2 

X 

o -35 

0 

xT 

X 

= + 

1.00 

<< 

(G'G) = 

- 3 

X 

o -35 


— — 

1.05 

<< 

(GC') = 

+ 4 

X 

0-35 

x see 45 0 

— + 

2.00 

u 

(C'C) = 





= + 

0.35 

(( 


The resultant upper chord stresses are the following: 

(C’G') = - 4 x 0.35 x tan 45 0 = — 1.40 tons, 

( G'M')— — 2 “ “ “ — 1.40 = — 2.10 “ 


SWAY BRA CING. 


375 


(GK) — + 4 x 0.35 x tan 45 0 = 4- 1.40 tons. 

(/CL) = + 2 “ “ “ + 1.40 = + 2.10 “ . 

The lower resultant web stresses are the following, remem¬ 
bering that the train* pressure is a moving load, and that 
D and O’ are the supporting points for the lower sway truss: 

(Q'R)=+ 6x0.30x^45° = + 2.56 tons. 

(QQ')=~ 6x0.30 -0.35 = - 2.15 “ 

(P'Q) = + 10x0.30 xsec 45° + 2 xo.63 x sec 45°= + 6.04 “ 
(PP) — — 10x0.30 —3x0.63+ 0.28 = — 4.61 “ 

(E'P) — + 15 x0.30x^45° +4x0.63 x^45°= + 9.96 “ 
(EE') = — 15x0.30 —5x0.63+ 0.28 = - 7.37 “ 

(DE) = + 21 xo.30 xsec 45 0 + 6x0.63 xsec 45°= + 14.31 u 

The + 0.28 ton, which is a release, is due to the fact that 
the half panel wind pressure against the floor system is 
added to 0.35 ton, and taken once too many times in each of 
the struts. 

The quantity 0.30 will be at once recognized as 2.10 + 7. 
The counters S'T and E’S are not required to resist wind 
stresses, but should never be omitted, in order that the 
general stiffness of the bridge may be increased ; their cross 
sections may be the same as that of QR . 

The lower resultant chord stresses are the following: 

(DE') = — (6x0.63 + 3x2.10) tan 45 0 = — 10.08 tons. 

(E'P') = — (4x0.63 + 2 x 2.10) “ — 10.08 — — 16.80 “ 

(P’S') — — (2 xo.63 + 2.10) 16.80 =— 20.16 “ 

(EP) — — (DE') + 2.10 = + 12.18 tons. 

(PQ) = — (E'P') + 2.10 = + 18.90 “ 

(QR) = — (P'S) + 2.10 = + 22.26 “ 

Although not a part of the lower chord of the truss under 
consideration, BE sustains the stress : 

(BE) = + 2.10 tons. 


* The train is taken as passing from right to left. 






376 WIND STRESSES AND BRACED BIERS. 

If the wind blows in the opposite direction to that assumed, 
the chord stresses which have been determined for C'N' will 
be found in CN, and vice versa. Precisely corresponding 
changes are to be made in the lower chords. The stresses in 
the sway struts would not be changed. That diagonal in 
each panel which is not stressed in the preceding instance, 
would sustain a tensile stress exactly equal to that already 
found in the other diagonal. 

It is therefore necessary to make calculations for but one 
direction of the wind. 

So far as equilibrium is concerned, in the preceding inves¬ 
tigation, there might be taken H" = — H and H' = o. In 
such a case BC would be subjected to a bending moment at 
F equal to — M» ; and the bending moment in CC\ at A, 
would be — while the stresses in FA, AC, and CF would 
be respectively — (FA'), — (A 1 C'), and — (CF'). For these 
reasons all parts of the portal should be built to sustain the 
stresses and moments which have been found when affected 
by opposite signs. 

It should be remembered that the parts EC, E'C', and CC T 
are subjected to combined direct stresses and bendings to 
the respective amounts that have been found. 

Those portions of the lower sway struts EE', PP', etc., 
extending from the windward rail to the lower chord BO 
(with the direction of the wind first assumed), are each 
subjected to a compressive stress, in addition to those already 
found, nearly equal to an amount to be determined in the 
following manner : Let N represent the number of panels 
in the sway truss, and n the number of any strut, from the 
farther end of the truss, counting the end itself zero, i. e., for 
PP', n will be 5. In the case taken N— 7. The amount 

desired will then be the panel train wind load multiplied by El 

JST 

added to the panel wind pressure against the floor system ; or in 
the example, 

n s s- 

2.10 x-1- 0.56 = 0.30 x n + 0.56. 


SWAY BRACING. 


377 

This compression in the struts arises from the fact that the 
wind pressure against train and floor system is not applied 
at panel points, but on the struts between their ends , and that 
a panel load of the former must be added to the ordinary 
stiut stress which exists with the head of the train at the 
strut considered. 

This involves, however, a very small error on the side of 
safety, since the pressure is divided between the two rails. 
Considering both directions of the wind, it will be seen that 
these struts are subjected to this amount of compression 
from end to end, in addition to the regular truss stresses. 

All the preceding wind stresses are to be combined with 
those due to the vertical loading, wherever they act in the 
same piece. 

If the portals are vertical, the stresses (BO) and (C'N'), due 
to the moment M lt will be zero ; also the span of the upper 
sway truss will be equal to that of the lower. No other 
changes will occur. 

If the bridge is a deck one, when possible, the ends of the 
chords should be secured directly to the piers or abutments, 
as no bending will then take place in the end posts. If this 
is not possible, the calculations will be precisely the same as 
those already indicated, with possible changes of signs in 
some of the stresses in the end posts or braces. In deck 
bridges, however, the wind pressure against floor system and 
train will be found in the upper chord. 

The method of treatment which has been exemplified is, 
therefore, perfectly general and sufficient for all cases. 

In deck bridges, tension sway braces (contained in planes 
normal to the trusses) are introduced, extending from either 
chord of one truss to the diagonally opposite one in the 
adjacent truss. So far as pure equilibrium is concerned, when 
horizontal sway trusses are present, these are superfluous; but 
they are very efficient in their influence on lateral stability. 
The actual stresses, in these members, in any given case, are 
indeterminate, but their greatest possible values are easily 
fixed. Let the total wind pressure exerted at a pair of oppo¬ 
site panel points in the two upper chords be represented by 


378 


WIND STRESSES AND BRACED PIERS. 


P\ and let a represent the angle between a horizontal line 
and the tension brace in question. Then the greatest possible 
stress which is required will be: T' = P' see a. 

This assumes that all the wind pressure is carried to the 
lower chords and resisted by the lower sway truss. 

In the case of vertical end posts, where the upper chord 
ends are not secured directly to piers or abutments, the 
stresses in the end lateral diagonals become perfectly deter¬ 
minate. In fact, in such a case, the braces AF, A'F', Figs, 
i and 2, become the diagonals in question. Let P represent 
half the total wind pressure in the two upper chords. The 
tensile stress in either one of these diagonals (if a retains its 
preceding signification) will then be; 

T — P l see a. 

Let P represent the total wind pressure against the bridge, 
and P" the total wind pressure against the train when it covers 
the whole span. Then let W and W represent the total 
weight of bridge and train respectively ; also let f be the 
coefficient of friction between the foot of end post and the 
supporting surface underneath. In order that neither truss 
shall possibly be moved bodily by the wind, with the bridge 
empty or covered by a train, there must exist the following 
relations : 


P<f(-J~ 2V ) : or > P+ P" </( 


/ VV f 



In the cases of through bridges, or of deck bridges with 
upper chords not secured to abutments, V 1 is to be found by 
applying the general form of Eq. (3) to both bridges and train. 

If the ends of the chords are secured to the piers or abut¬ 
ments, the resistances of these fastenings will take the place 
of the frictional resistances. 

One other effect of the wind pressure against the train 
remains to be noticed. The normal action of this pressure 
will not permit the train’s weight to be distributed between 
the two chords which carry it, according to the law of the 



SWAY BRACING. 


379 


lever. Let p" represent a panel wind pressure against the 
train, and let h represent the height of its centre of ac¬ 
tion above the points of support. Also let b represent the 
horizontal distance between centres of trusses; then 


t 



( 4 ), 


will be the amount of load which is transferred from the 
windward to the leeward truss. In other words, the panel 
leeward load will exceed the panel windward one by 21 . If, 
therefore, w is a panel moving load, the action of the wind 
will cause all moving load truss stresses to be increased by an 
amount found by multiplying the stresses, determined with¬ 
out regard to the wind, by— i . Also, if s is the distance 

(normal) between two adjacent parallel stringers, the increase 
of load on one, and decrease of that on the other, will be: 




Eq. (5) gives the variation of load on the floor beam also. 
Without essential error, h may be measured downward from 
the centre of the body of the car. 

Specifications sometimes require calculations to be made 
with an unloaded bridge. In such a case the methods are 
precisely the same as the preceding, with the train wind 
pressure omitted. 


Art. 82.—Transverse Bracing for Transferring Wind Stresses from One 
Chord to Another—Concentrated Reaction. 

In the preceding Article it has been supposed that the wind 
pressure is resisted by sway trusses in the horizontal planes of 
both upper and lower chord. It may sometimes be desirable 
to transfer all the wind pressure to the lower chord, or to 
the upper. 






WIND STRESSES AND BRACED PIERS. 


380 

The section of a through truss bridge, in which it is 
desired to carry all the wind pressure to the lower chord, is 

represented in Fig. 1. AC and ND are 
posts directly opposite to each other in 
the two trusses. AN and OB are lateral 
struts , while AO and BN are lateral ties. 
Let the wind be supposed to blow from 
right to left, as shown by the arrow. 
According to the principles of the pre¬ 
ceding Article, in consequence of its 
direction the wind will relieve the truss 
AC of a part of the weight which it 
carries, and add the same amount to that carried by the 
truss DN. 

If the direction of the wind were reversed, the truss DN 
would be relieved, and AC would receive the increase of 
loading. 

Let this relief (or increase) of truss load, per panel, be de¬ 
noted by w ; it will act as though hung from B. 

The following notation, also, will be used: 

AB = d = ON. BC — a — OD. 

DC = AN = b. 

F — total wind pressure, per panel (for one truss), on 

i(AB + BC). 

F' — total wind pressure, per panel (for one truss), on 

\AB. 



With the assumed direction of the wind, the tie AO will 
not be stressed. As usual, the plus sign ( + ) will indicate 
tension, while the minus (—) sign will indicate compression. 

In this article the total horizontal reaction, equal to 
2 (F + F'), will be taken as concentrated at D. 












TRANSVERSE BRACING. 


381 


There will then result: 
Relief in truss AC — w 

Compression in AN — 
Tension in BN — 
Compression in BO — 


_ fF' (a + d) + Fa> 



- (AN) = 




+ (BN) = + w sec ABN. (3). 

— (BO) — — (F + w tan ABN ) 


= — (f + 2F' + (F + F')~) 

= F-2{F+F>) (^-~). ( 4 ). 
Compression in ND = — (ND) = — w. . . (5). 


The horizontal force 2 (F 4- F') acts toward C, at D, pro¬ 
ducing a bending in DN which has its greatest moment M 
at 0 . Hence: 


M = 2(F + F')a .(6). 

If K is the greatest intensity of compressive stress (due to 
flexure) in the cross section of the post DN , at 0 , d x the 
greatest distance of any compressed fibre from the neutral 
axis of the cross section, and I the moment of inertia of the 
section about an axis passing through its centre of gravity 
and lying in the plane of the truss ; then, by the well-known 
formula— 


K = 


d x M 
1 



At 0 there will then exist the intensity of compression : 



; in which q is the area of cross section 


of 







382 


WIND STRESSES AND BRACED PIERS 


f w \ 

the column. The intensity of compression — — T Kj is 

in addition to the regular truss stresses arising from vertical 
and wind loads. 

If, for lack of head room, a flanged beam only is used, 
as shown in Fig. 2, instead of the lateral bracing of Fig. 
i, then that beam will be subjected to combined compres¬ 
sion and bending. Let + F represent the total wind pressure, 

per panel, for both trusses , on \AB, and let w 
represent the release of weight in AB and in¬ 
crease in CD. 

Also let 

AB — a , and BD — b. |Ais to be taken as 
W B applied at A and w, at the same point. Equal 
and opposite forces are also to be supposed 
to act at D. The moment 

M — Fa — wb, exists at C and gives: 


raA 


W = 


Fa_ 

b 


With the direction of wind shown by the arrow, the bend¬ 
ing caused by w will increase uniformly from nothing at A to 


M — wb 


at C. The bending moment, therefore, to be resisted by this 
beam AC, and by the joints between it and the chords A and C, 
is : 

M — Fa — wb .(8). 


The direct compression in A C is : 

(AC)=-iF .( 9 ). 

Hence, if q is the area of cross section of the beam, and if 
K, /, and d x retain the same general signification as in Eq. (7), 
the greatest intensity of compression in the beam (at its 
ends) will be: 












TRANSVERSE BRACING. 


383 




The direct compression in CD is 

(CD) = — w. 




The bending moment in CD at C is: 

M = Fa .(12). 


The greatest compressive intensity is found at once by 
Eq. (10), after writing w for \F, and giving to 
the remaining notation its general signification. D 
The two preceding cases are those of through 
trusses. In the case of a deck truss the lateral 
bracing is of much more simple character ; it is 
shown in Fig. 3. At C and A are the two lower 
chords. CA is a lateral strut, while BC and c 
DA are lateral ties. No parts are subjected to 
bending. 

If F is the panel wind pressure (for both trusses) acting 


along AC, there will result : 

(CA)=-\F. .(13). 

(BA) = — w .(14). 


(BC) = -f VP 2 + w 2 .(15). 



The horizontal component of (BC) is equal to F, and 
acts at B. Thus all wind pressure is carried to the upper 
chord. 

The compression (BA) is in addition to the regular truss 
stresses induced by the vertical and wind loads. 

By these methods all the wind pressure may be carried to 
either chord. The truss stresses of the sway truss in the 
horizontal plane of that chord have already been found in 
the preceding Article, or rather, the methods for finding 
















384 


WIND STRESSES AND BRACED PIERS. 


them and the effect of w on the stresses in the vertical 
trusses have there been completely given. 

The wind has been taken in one direction only; with the 
other direction, opposite but symmetrically located parts 
would be stressed by the amounts found. 


Art. 83.—Transverse Bracing with Distributed Reactions. 

In the preceding articles it has been assumed that the hori¬ 
zontal reactions of the wind pressure were concentrated at the 
extremity (top or bottom, as the case may be) of one post in 
the tranverse panel considered. This assumption, however, 
may not be admitted; or some other may be substituted in 
its place. 

Let Fig. i represent a transverse panel, with the wind 
blowing in the direction shown by the arrow. 

As before, the following notation will be used : 

AB — ON — d. BC ~ OD = a. 

DC = AN = b. 


F =total wind pressure, per panel, for one truss, on \{AB + BC). 

p' — “ “ “ “ 

w = relief of load in truss AC. 


u 


a 


u 


“ \AB. 



Instead of concentrating the entire hori¬ 
zontal reaction at D, if ?i is a quantity less 
than unity, there will be assumed: 

Horizontal reaction at D — 2 n (F + F'). 

11 C = 2(i— n)(F + F'). 

D ifij-jjji '||jU|[fc The wind pressures on \BC — \OD act 

Fig z directly at C and D in the horizontal sway 

truss, and, consequently, will be omitted 
from consideration. 

As in the preceding article : 


rr ._ 2 (F(a + d)+Fa) 


■ (!)• 












TRANSVERSE BRACING. 


3^5 


Taking moments about B: 


(AN) = - 


F’+ 2(i-n) (F + F 



Taking moments about N: 



(OB) = 


2n(F + F') (a + d) 

_ d 




Taking moments about the intersection of AN and OB at 
the distance infinity (oo) from the figure: 

(BN) oo cos ABN = + [w oo + 2 n(F + B')a^\ ; 


(BN) = .+ w sec ABN = l l.sec ABN. (4). 


The stress in BN> therefore, remains the same whatever 
may be the assumptions in regard to the horizontal reactions. 

The bending moment at O, about an axis lying in the plane 
of the vertical truss, will be : 


M = 2n(F +F')a .(5). 

Since the windward truss is always relieved of a part of its 
weight the bending moment 2(1 —n)(F + F')a, at B, will 
seldom or never be needed. 

The value of M, from Eq. (5), put in Eq. (7) of the pre¬ 
ceding Article, and in the expression following that equation, 
will enable the greatest compressive intensity in the post to 
be found. 

If the transverse panel, Fig. 1, represents the portal of a 
bridge, the distances AB and BC, or d and a , represent in¬ 
clined distances in the plane of the portal. F' will (or may) 
then include , also , the reaction of the horizontal sway truss in 
the plane of AN\ while F will include the reaction of the 
horizontal sway truss in the plane of OB, if there is such a 
sway truss. 

35 . 








386 


WIND STRESSES AND BRACED PIERS. 


If n — as is sometimes assumed : 



( 0 - > 
• ( 7 )- 


If n = i in the formulae of this Article, those of the cor. 
responding cases in the preceding Article at once follow. 

In Fig. 2 let the notation be as follows: 

AB = CD = a. BD = CA = b. 


Total wind pressure for both trusses, per panel, along 


AC=F 

Horizontal reaction at D = nF. . . 

“ “ “ B={i-n)F. 


( 8 ). 

( 9 )- 


If Fig. 2 represents a portal, F will or may include the 
reaction of a horizontal sway truss. 

The bending moment on both DC and CA f 
a t C, also on the joint at the same point, is : 






Mi — nFa .(io). 


Fig. 2. 


This is the greatest bending in DC and CA 
of that kind which produces compression in the 
lower flange of the beam CA. 

The relief of panel load in the truss AB and increase of 
that in CD is ; 


w = 


Fa 

b 


(io). 


Let x represent any variable portion of CA; then the 
bending moment at any point of CA is: 
















STRESSES IN BRACED PIERS. 


387 


Fa 

M —n Fa — wx = n Fa - x. . . (11). 

For the point or joint A, x becomes equal to b, while the 
expression for the moment is : 

M'\ — — (1 — n) Fa .(12). 

This is the greatest bending of the kind opposite to M x , in 
CA. It is also the greatest bending in AB. The con¬ 
nections at C must resist the moment M x , while those at A 
must resist M\. 

The direct compression in CA is %F. 

“ “ “ “ CD “ w. 

If n = i; M x = —M\ = \Fa .(13). 

These various bending moments, substituted in Eq. (10) of 
the preceding Article for M, will enable the greatest intensi¬ 
ties of stress in the members CA, CD, and AB to be at once 
found. 


Art. 84.—Stresses in Braced Piers. 

The general treatment of stresses in braced piers may be 
exemplified by that of a single “ bent ” represented by a 
skeleton diagram in Fig. 1, in which the horizonal web mem¬ 
bers are compressive ones. The plane of the “ bent ” is verti¬ 
cal and normal to the centre line of the truss whose end rests 
upon it; or if the track is curved, this plane is normal to it. 
The bent shown in Fig. 1 may be considered one of a pair, in 
parallel planes, which, being braced together, compose the 
complete braced pier. The dotted rectangle AMNB repre¬ 
sents a skeleton section of the truss supported by the piers, 
the upper chords of which rest upon the top of the pier at A 
and B. A skeleton section of the train is also shown. 

The direction of the wind is supposed to be shown by the 
arrows a, normal to the track at the top of the pier. If the 



3 88 


WIND STRESSES AND BRACED PIERS . 


trusses are loaded with a train, the wind pressure against 
them and the train will be carried to the top of the piers in 

the manner shown in Art. 81. The 
wind will also act against the pier 
itself. 

Let the train be supposed to cover 
the whole of the two spans adjacent 
to the top of the bent (in all ordi¬ 
nary cases one of these spans will be 
the distance between two adjacent 
bents); then let H represent half 
the total pressure against trusses, 
and Pf half that on the train cover¬ 
ing the two spans. 

The height of the centre of action 
of Pf above AB, Fig. i, is k. Also 
let b — AB. The pressure Pf will 
F ' decrease the train reaction at A and 

increase that at B by the amount: 
Fig. i. J 



Vi = 


p;'h 


(i) 


Let h! represent the vertical distance of the centre of ac¬ 
tion of H from the horizontal line AB. 

The wind pressure on the truss AMNB will cause an in¬ 
crease of truss reaction at A , and an equal decrease of that 
at B , which will be denoted by V, and its value will be: 


V = 


Hit 1 
b ’ 


consequently if 



Hh' 
b ’ 


the total horizontal force to be taken as acting at A, and with 
the wind, will be ( H + Pf — 21 ' tan a) added to the wind 
pressure acting directly at A. In Fig. 2, ^represents this 


















STRESSES IN BRACED PIERS. 389 


force, laid down to any desired scale. The small segments 
measured to the right of d represent 
the panel wind pressures against the 
pier at the points C , E , G, and K, 
while those shown on the left of c rep¬ 
resent the panel pressures at D, B , 

E, H , and L . The panel pressures at 
A, B, K, and L are half those at the 
other points. 

Let W represent the total weight 
of adjacent trusses and moving load 
resting at the top of the pier. 

Let W x represent the panel weight 
of the pier itself resting at the points 
C, E , G, D, F, H: £ W 1 will be taken 
as applied at the points A , B, K, and 
L; then the resultant reactions at A 
and B, with the wand blowing, will 
be, respectively, 



W ,, , W „ 

- t and- V t. 

2 2 



It has been implicitly supposed that two equal and opposite 
forces, equal in magnitude and parallel to PG, act along AB. 
One of these forms, with PG itself, the couple PGh ; the other 
is the wind pressure which, combined with the half panel press¬ 
ure at A, and (H— 2t' tan <*), is represented by cd in -Fig. 2. 

The quantity t' is the force of a couple whose lever arm is 
b. One force t' is therefore supposed to act at A, and the 
other at B. V x will be considered larger than V; hence t 
will act upward at A and downward at B. If a is the angle 
between AK or BL and a vertical line, the t' at B will cause 
a compression in AB equal to t' tan a, while the t' at A will 
pull to the left by the same amount. Consequently the force 
2 t' tan a will act on the point A and toward the left. 

In the diagrams and in the equations which follow, positive 
and negative signs indicate tensile and compressive stresses, 
respectively. 







390 


WIND STRESSES AND BRACED PIERS. 


The stresses due to vertical loads at A and B, and tht 
other panel points, will be the following: 



(CD)" = — W t tan a. 

(. EF) n = - “ 

(GH)" = “ 

(KL)"= + f + 7 -p) tan a. 


The difference between the horizontal component in HL 
and (KL)" is 21 ' tan a, and it acts towards the right. 

The stresses caused by the horizontal wind pressure acting 
through A, B, C } D, etc., are shown in Fig. 2, as has already 
been noticed. The diagonals sloping similarly to AD are as¬ 
sumed not to be stressed. The diagram explains itself. 

The resultant stresses, finally, are to be found by combin- 





STRESSES IN EE A CEB PIERS . 


391 

ing the results of the diagram in Fig. 2 with those expressed 
by the equations already wiitten. dhey are the following! 


(AC) = - 1 

(f- 

, W\ 
t + 2) 

o'. 

(CE) = - ( 


“ + ?) 

u 

+ ((FA). 

(EG) = -1 

f « 

" +¥) 

u 

+ (EG). 

(GK) = - 1 

(« 

■ +*F) 

u 

+ (GK). 

(BD) = - ( 

f W 

\ 2 

' + 7 ) 

sec 

a - (BD). 

(df) = - ( 


“ + iK' 

2 / 

) “ 

- (DF)- 

(EH) = - ( 

f u 

.. + 5 

2 / 

) “ 

~ (FH). 

( 77 Z) = -( 

f u 

„ + 7 w ;n 

2 y 

) “ 

- (HE). 


(AB) = -i(W + W) tan a - (AB). 

(CD) = - W, tan a - (CD). 

(EF) = - “ “ - (EF). 

(GH) = - “ “ - (GH). 

(KL) = + (y - t’ + tan a - (KL). 

It is not necessary to reproduce the stresses in the oblique 
web members, since they can be scaled directly from Fig. 2. 

All the stresses may be checked by the method of 
moments in the usual manner, and such checks should always 
be applied. 

The two reactions R and R' are the following: 

R = \(W - 2t f + 8 W x ) + R t . 

R ' = i(W + 2t ' + 8 W x ) + R t \ 










392 


WIND STRESSES AND BRACED PIERS. 


It is to be remembered that R( is to be taken as positive 
in these expressions ; also that RJ= — R h as shown in Fig. 2. 

The lateral force F x to be resisted at the foot of the bent 
by friction or some special device, is the total wind pressure 
against the train, truss, and bent. 

If f is the coefficient of friction at K and Z, Fig. I, the 
lateral resistance of friction offered at K is f'R, and that 
at L,f'R r . It is supposed that both the reactions R and 
R' are upward , also that both coefficients of friction are the 
same. 

The expression for (KL) has been written on the assump¬ 
tion that all frictional resistance is exerted at L. Strictly, 
however, the stress in KL may be taken as: 

(KL), = (KL) - f’R ; 


always supposing, numerically, ( KL ) > f'R . 

The circumstances of particular cases frequently require 
calculations to be made with the structure free of moving 
load, as well as covered with it. In such a case it is only 
necessary to put for W, in the preceding operations, the 
weight of trusses only. 

The wind has been taken in but one direction only, though 
the pier is to be designed for both directions, since it is only 
necessary in the resultant stresses to change the letters B , D , 
F f Hy L, to Ay C, E, Gy K , and vice versa. 

If MNy Fig. i, should coincide with AB, or if the truss 
should rest upon the top of the pier, it would only be 
necessary to take t' — V t + V, remembering that h is the 
distance (vertical) from the centre of P” to the top of the 
pier. 

It should be stated that 2 t' tan a may be treated as a 
single force acting toward the left and along AB, Fig. I. It 
will then give rise to the diagram in Fig. 3, which shows all 
the stresses produced by its action. In that Fig. ad repre¬ 
sents 2 t' tan a. In such a treatment of the question, cd t 



STRESSES IN BE A CEE PIERS. 


393 


2(AB)~ 


Fig. 2, would represent H 4- P added to the half panel 
pressure at A. The resultant stresses would then be found 
by combining the results of the two diagrams with those of 
the equations. All the results of these 
two methods will not agree ; the latter a “^ 
will give the greatest. This ambiguity 
cannot be avoided, for it results from 
the fact that the pier cannot be so 
divided as to sever these members 
only. 

Mr. J. A. Powers, C. E., has called the 
attention of the writer to the fact that 
the web members of a braced pier carry¬ 
ing a double track railway, similar to 
that shown in Fig. 4, will receive their 
greatest stresses with the windward track 
only loaded. 

The vertical member GH may be sup¬ 
posed to carry its proper proportion of 
the load which rests on each track. This FlG - 3 * 

supposition, however, does not affect the statement made 

above. 

Let the wind have the direc¬ 
tion shown by the arrow, and let 
W, as before, represent the fixed 
and moving weight resting on 
GB, while w' is that part of W 
which is carried to B. If GH 
acts * 

QD 




w = 


W. 


QF 

If GH does not act: 


w 


- c ~w 


DF 

In the latter case, the beam AB will carry -pp W to A, 



























394 


WIND STRESSES AND BRACED PIERS . 


If the angle FBN — CAM — a, the force 

h — w tan a — tan a 

will act along AB as an unbalanced horizontal one. If GH 
acts, Wi tan a becomes equal to zero. 

Then in the preceding investigation, there is to be put, 
(.H + h) for H, while w' is to be taken as acting vertically 
down at B, and or o (as the case may be) at A. The 
preceding methods and diagrams remain exactly the same as 
before. 

W 

In the formulae, however, or o is to be put for the — 

at A, Fig. i, and w’ for that at B in the same figure. Noth¬ 
ing else is changed. 

If W rests on AG and GB at the same time, a horizontal 
force equal and opposite to h is developed at A, Fig. 4. 
Hence h will be balanced and disappear. 

If W rests on AG alone, with the direction of the wind 
remaining the same, h will change its direction, thus giving 
much smaller web stresses than those existing with W on 
GB alone. 

If, for any reason, the load on a single track pier does not 
rest over its centre, h will have a definite value, and the above 
considerations must govern the determination of the web 
stresses. This condition may exist if it becomes necessary 
to place braced piers under a single track railway curve. 

The stresses caused by the traction, or pull, of the loco¬ 
motive, in the members of a braced pier, are simple in char¬ 
acter and easily determined. 

In such a case, the pier is simply a cantilever with the 
traction, or pull, as a single force acting at its extremity. 
The traction acts along the line of the rails, and the length 
of the cantilever is the height of the pier. The stresses thus 
determined are to be combined with those already found. 


COMPLETE DESIGN OF A RAILWAY BRIDGE. 395 


Art. 85.—Complete Design of a Railway Bridge. 

The main sections and details of this design shown on Pis. 
XI. and XII. are based on the following general specifica¬ 
tions. 

The span length, depth of truss, panel division, moving 
load, fixed loads, and stresses resulting from the preceding, 
shall be as determined in Art. n. 

The clear width between trusses shall be 14 feet. 

A wind load of 150 pounds per lin. ft. of span shall be 
taken for the upper chord, and the same amount for the 
lower, and shall be treated as a fixed load in each chord. In 
addition to this fixed load, 300 pounds per lin. ft. of span 
shall be taken as a moving wind load for the lower chord, 
since the train passes along the latter. 


The greatest allowed tensile stresses under the preceding loads 
shall be: 

For lower chord eye-bars.10,000 lbs. per sq. in. 

“ main brace eye-bars nearest end of span .... 10,000 “ “ “ 

“ first counter-brace.7,333 ** “ “ “ 

“ vertical adjacent to end of span.7,333 “ “ “ “ 

Other tension braces to be proportioned according to loca¬ 
tion between. 7,333 and 10,000 “ “ “ “ 

For plate hangers on floor-beams (net section) .... 8,000 “ “ “ “ 

“ bottom flanges of floor-beams and stringers (net section) 8,000 “ “ “ “ 

'* lateral braces . , .12,000 “ “ “ 

The greatest allowed compressive stresses shall be : 


For upper chord and end posts: 


Flat ends. 



7,800 


T 

I H -1 

5o,ooor 


Pin ends. 



7,800 

ZH 

30,ooor 2 



For intermediate posts at centre of span, a reduction of 20 
per cent, shall be made from the preceding values, and all 
other intermediate post stresses shall be proportioned accord¬ 
ing to location, between end and centre values. 











39 ^ COMPLETE DESIGN OF A RAILWAY BRIDGE. 


In the preceding column formulae, is pounds per 

square inch ; “/,” length; and “ r ” the radius of gyration of 
normal section in direction of failure, and in the same unit 

as “ ir 

For lateral compression braces the above values may be 
increased 20 per cent. 

For top flanges of stringers and floor-beams the greatest 
compressive stress shall be 7,000 pounds per square inch of 
gross section. 

The greatest mean bearing intensity of pressure between pins 
and pin-holes or rivets and rivet-holes, shall be 12,000 pounds 
per square inch. In stringers and floor-beams and their con¬ 
nections with each other or with the trusses, this value shall 
be reduced 25 per cent. 

The greatest shearing i?itensity in rivets or pins shall be 
7,500 pounds per square inch, and in stringers and floor- 
beams and their connections with each other and the trusses, 
this amount shall be reduced 25 per cent. 

The greatest bending stress in the extreme fibres of wrought- 
iron pins shall be 15,000 pounds per square inch, and the cen¬ 
tres of pressure shall be taken at the centres of the bearing 
surfaces. 

No cast-iron whatever shall be permitted in any part of the 
structure, and all parts shall be accessible for inspection and 
painting. 

The unsupported width of any plate in compression shall 
not exceed thirty times its thickness. 

The pitch of rivets in compression members shall not ex¬ 
ceed sixteen times the thickness of the thinnest plate through 
which the rivets pass. 

An initial stress of 5,000 pounds shall be added to that 
produced by the vertical loading in all adjustable tension 
members. 

These meagre specifications are sufficient for the design 
when it is premised that all details of construction, such as 
eye-bar heads, connections, etc., shall be consistent with the 
best engineering practice. 


COMPLETE DESIGN OF A RAILWAY BRIDGE. Z 97 


Although it is customary to add the total stress of adjust¬ 
ment to that caused by the vertical loading, in adjustable 
tension members, the practice is not strictly correct. Just 
what part of the initial stress should be added is not exactly 
determinate, but it is certainly not the whole. For this rea¬ 
son the apparently small value of 5,000 pounds has been taken. 

As the widths of the compression members depend, to 
some extent, on the thickness of the tension members, and 
as the design of the latter is of the greatest simplicity, it 
conduces to the greatest convenience to begin with them. 
All eye-bars should be as thin as considerations of resist¬ 
ance will permit, as pin bending will then be reduced to a 
minimum. 

By taking the stresses from Fig. 1 of PI. II., and subjecting 
them to the preceding specifications, the following sections 
are obtained : 


Brace. 

Total stress. 

Allowed stress. 


Sections. 


2 

45.553 

lbs. 

7,333 

lbs. 

per sq. 

in. 

2-4" X J" 

Bars. 

3 

167,800 

<< 

10,000 

a 

a 

11 

11 

2-6 X if 

a 

5 

119,817 

n 

9,340 

It 

11 

11 

ii 

2 — 6 X if 

11 

7 

77.445 

it 

8,670 

a 

11 

a 

a 

2-5 X f 

n 

9 

47,654 

it 

8,000 

a 

11 

ii 

ii 

2-2''* 0 

11 

10 

21,560 

n 

7,333 

11 

11 

ii 

a 

2-1* “ 

11 

Lower chord. 









I = 2 

131,520 

it 

10,000 

ii 

11 

it 

ti 

2-5"xi"| 

11 

3 

225,424 

n 

10,000 

u 

11 

11 

ii 

4 - 5 " xi* 

11 

4 

289,238 

n 

10,000 

11 

11 

11 

a 

6- 5" xi 

a 

5 

322,220 

It 

10,000 

11 

11 

ti 

11 

6-5 xi T V 

11 


There should be as little diversity in widths of bars as pos¬ 
sible, but varying thicknesses within standard limits are easily 
produced in the mill. Again, a small number of large bars is 
cheaper to produce than a large number of small bars, on 
account of the smaller number of pieces. Hence the aim 
should be to produce large pieces, though not too heavy to 
be handled conveniently in the shop. 

Before designing the pins the intermediate post sections 


39 ^ COMPLETE DESIGN OF A RAILWAY BRIDGE. 









r—T - 

I 

1 

I I 

O 



~ IV 


T 

1 


C"—s 



should be determined. These posts are secured to pins at 
each end, and although they are constrained, by some extent 
in the vertical plane of the pin axis, it is only slightly so, and 
they should be considered columns, with pin ends in all direc¬ 
tions. They will each be built of two channels laced in the 
usual manner. The depth of the channel is an important 
matter, but the length of no column in a truss should exceed 
forty times its least diameter, and in the present case the 

depth will be taken at ten inches. The 
channels will be placed as shown in the 
figure with a clear separation of ten inches. 

Pin plates will be riveted to the flanges 
of the channels at each end, through which 
the pin will pass, leaving the axis of the 
latter parallel to the channel webs and nor¬ 
mal to the planes of the trusses. The least 
radius of the post section will be parallel to the pin axis and 
will be the same as that of one channel about an axis normal 
to its web, or about 3.9 inches. The length of the post 
between pin centres is 27 feet, or 324 inches. But in the 
plane normal to the truss the column is shortened six feet 
by the transverse bracing as shown in Fig. 16 of PI. XII. 
Hence, in the plane of the pin axis / -4- r ~ 252 -4- 3.9 = 65 ; 
and in the plane normal to the preceding r = 5.8 .*. / -4- r = 
324 - 4 - 5.8 = 56. As the post is considered with pin ends in 
all directions the first value of l -4- r will be used. 

Eq. (1) then gives for a post at the end p — 6,840; and for 
one at the centre 0.8 x 6,840 = 5,472. Now since (6,840 — 
5,472)- 4 - 3 = 456; 


Fig. 1. 


For vertical brace 4. ./ = 6,840 — 456 ~ 6,384 lbs. per sq. in. 

“ “ “ 6. .p = 6,384 - 456 = 5,928 “ “ “ “ 

“ “ “ 8.^=5,928-456=5,472 “ “ “ “ 

The initial stresses in the counters intersecting at the top 
of vertical brace 8 increase the stresses in that member 8,000 
pounds. The preceding quantities then give the following 
results: 











COMPLETE DESIGN OF A RAILWAY BRIDGE. 399 


Total stress. Allowed stress. Member. 

Vertical 4 .. .99,819 pounds... .6,384 pounds. .. .2 — 10 ' 79 lb. channels. 

6-66,193 “ 5,928 “ 2 — 10” 56 “ 

“ S-42,611 “ 5,472 “ 2 - 10" 48 “ 

The last sectional area shows a material excess over that 
required, but a 48-lb. channel is about the lightest rolled, and 
this excess is usually found in the centre posts of trusses. 

The lacing on these posts will be 2 \ x ^ placed at an 
angle of about 6o° with the post axis. 

In order to determine the lower chord pin bending some 

1—2 _ 65800 _ 

65300 1 ■ 3 



CO 


_Centre_i_Line_ 

45553 

Fig. 2. 

diameter of pin must be assumed, for the moment at the 
centre of the pin will depend partially on the thickness of the 
pin plates, which bear against the pins. A diameter of 4! 
inches will be taken ; hence each inch in length of the pin 
will take 4,375 x 12,000 = 52,500 pounds. It will be seen 
hereafter that the floor-beams will be riveted into the posts 
below the pins in such a manner that the pin plates will not 
only carry the column pressures to the pin, but the floor- 
beam loads also. Hence, in determining the thicknesses of 
bearing areas in the pins, these two loads and the pressures 


65800 



Centre _Line 


Fig. 3. 

due to initial stresses in the counter rods must be added. 
The vertical brace 8 is the only post subject to the initial 
stresses in the counters and the vertical component of each 
counter at its top is 4,000 pounds, or 8,000 for the two. 






















400 COMPLETE DESIGN OF A RAILWAY BRIDGE. 


The total lower chord load has already been found to be 45,- 
553 pounds in the case of brace 2. This maximum lower chord 
panel load will not usually occur with the greatest post stress, 
but all possible cases are covered by combining the two. The 
bearing thickness at each side of each post is thus found to be: 

For brace 4.(99,819 + 45 > 553 ) 5 2 > 5 °° x 2 = 1.4 inches. 

“ “ 6.(66,193 + 45,553) -*-52,500 “ =1.06 “ 

“ “ 8.(34,611 + 45,553 + 8,000) -f- 52,500 “ =0.84 “ 


By regarding the principles affecting pin bending as de¬ 
veloped in Art. 74, it will be found that the arrangements of 



Fig. 4. 


lower chord eye-bars and braces shown in Figs. 2, 3, 4 
and 5 will reduce the lower chord pin bending to the least 
amounts possible. 

The values of these least pin moments for the principal 


4S200 


4—5 


a' - r— . - 


53700 



Fig. 5. 


sections are found to be as follows, and they can be verified 
by remembering the thicknesses of the eye-bars (already de¬ 
termined) and the fact that a play of one-eighth of an inch is 
allowed between each contiguous pair of heads. 

Joint 1—2. 

The section of plate hanger at the end of the floor-beam is 
shown shaded, and the distance between its centre and that 






































COMPLETE DESIGN OF A RAILWAY BRIDGE . 401 


of either of vertical braces 2 is 3 inches. Taking moments 
about the centre of the plate-hanger: 

Moment about Vert, axis.65,800 x 1.5 = 98,700 in. lbs. 

“ Hor. “ .22,800 x 3.0— 68,400“ “ 

Hence resultant moment — 4/ (98,70o) 2 + (6S400) 2 = 120,000 “ “ . (2). 

Joint 2—3. 

The bearing area of post 4 on the pin is shown shaded. In 
the remaining cases it will be assumed that the adjacent 
lower chord panel stresses take their greatest values together, 
in accordance with which assumption the eye-bars will be 
stressed for this joint as shown in Fig. 3. The force 78,400 
pounds is the corresponding stress in one eye-bar of brace 3. 
The tangent of the inclination of the latter to a horizontal 
line is 1.32, hence the vertical component of brace 3 (one eye- 
bar) is (112,800 — 65,800) 1.32 = 62,040 pounds. 

The moments about vertical axes are : 


About a .65,800 x 2.625 — 56,400 x 1.25 = 102,225 in. lbs. . (3). 

“ b . 65,800 x 4.0 — 112,800 x 2.0 = 37,600 “ “ 


The moment about a horizontal axis through c is: 

62,040 x 1.5 = 93,060 in. lbs. 

The vertical moment about c is the same as that about b, 
hence the resultant moment about c is: 

4/ (37,600) 2 + (93,060) 2 = 100,360 in. lbs. . . .... (4). 


Joint 3—4. 

By the preceding method the resultant moment of the in¬ 
clined brace was resolved into vertical and horizontal com¬ 
ponents ; in this and the following cases, however, the 
resultant moment itself will be taken. As before, the post 
bearing is shaded in Fig. 4. The head of the counter rod c 0 
is assumed to be one inch thick. The secant of the inclina¬ 
tion of the inclined bar to a horizontal line is 1.66. Hence 
the inclined stress is (3 x 48.200 — 2 x 56,400) 1.66= 53> 1 

pounds. The vertical moments are then as follows: 

26 








402 COMPLETE DESIGN OF A RAILWAY BRIDGE. 


About ci . . . 48,200 x 1.2 = 57*840 in. lbs. 

“ a . . (112,800 — 96,400) x 2.94 = 48,216 “ 

“ b ... . 48,200 X 3.5 — 16,400 x 6.44 = 63,100 “ 

The inclined moment about b is; 

53,166 x 2.31 — 122,800 in. lbs. 

In Fig. 6 ab is normal to the inclined brace, and represents 
122,800 inch-pounds by scale, while be is vertical, and repre¬ 
sents 63,100 inch-pounds. 

Hence the resultant moment about b is represented by ac y 
and has the value: 

R — 97,500 inch pounds.(5). 

Joint 4—5. 

The thickness of the head of the centre rod c is 1.5 inches; 
and the inclined eye-bar stress of brace 7 is 

(3 x 53,700 — 3 x 48,200) 1.66 = 27,500 pounds. 

The vertical moments are as follows: 

About d 48,200 x 1,155 

“ a 2 (53,700 — 48,200) x 2.875 
“ b 2 (53.700 — 48,200) x 4.03 — 48,200 x 1.155 
. “ d 3 (53,700 — 48,200) x 3.7 — 11,340 

t 

The inclined moment about d is 

27,500 x 2.6 = 71,500 in. lbs. 

Hence the resultant moment about d is, by Fig. 7: 

R — 57,ooo inch-pounds.(7). 

In addition to the preceding, the moments of the greatest 
vertical components of the inclined eye-bar stresses about the 
centres of the post bearings should be examined. It is here 
unnecessary to go into all these in detail. The greatest oc¬ 
curs at joint 3—4. The vertical component of the maximum 


= 55,671 in. lbs. (6). 

- 31,625 “ “ 

= — 11,340 “ “ 

= 49,700 “ “ 



COMPLETE DESIGN OF A RAILWAY BRIDGE. 403 


stress is (119,817 2) 0.8 = 47,930 lbs. Hence the moment 

in question is: 


47,930 x 2.31 = 110,700 in. lbs. . . . (8). 

The preceding results show that while it is quite unneces¬ 
sary to take moments at the centre of all bearings, a thor¬ 
ough examination of the lower chord joints must be made in 
order to find the greatest moments. 

Eq. (2) gives the greatest resultant moment of 120,000 
inch-pounds. Hence a wrought-iron pin 4,375 inches in diam¬ 
eter will be sufficient to meet the requirements of the speci¬ 
fications. But since the bars in braces 3 and 5 are 6 inches 
wide, and since the eye-bar heads are no thicker than the 
bodies of bars, the requirement of 12,000 pounds per square 
inch bearing pressure against pins cannot be met by a less 
diameter than 5 inches in the case of brace 3. For a reason 
that will appear hereafter, the diameter of the large pins will 
be taken at 5| inches. 



In the cases of braces 7 and 9 a much smaller pin may be 
used, and while it is not economy in the shop to have a large 
number of pin diameters, two, or even three, are not too many 
for a span of this length. The cosine of the inclination of 
brace 7 to a horizontal line is 0.61, hence the horizontal 
component of the greatest stress in one of its eye-bars is 
(77,445 -r- 2) x 0.61 = 23,620 pounds. It will be seen here¬ 
after that the side plates of the upper chord panels 3 and 4 
will be 0.5 inch thick, and since 4j\ x 12,000 ~ 2 = 25,000, 





4°4 COMPLETE DESIGN OF A RAILWAY BRIDGE. 


it appears that with a pin diameter of 4 T 3 ¥ inches no thicken¬ 
ing plates at pin-holes D , E, E, and M will be needed. It 
will be found that the thickness of bearing plates at the top 
of brace 6 must be J-J- inch, and the clearance at each side of 
eye-bar head (between J-inch side plate of chord and pin plate 
of post) will be about T 3 g inch. The vertical component of 
eye-bar stress (for brace 7) will be (77,445 2) x 0.8 = 30,980 

pounds. Hence: 

Pin moment at centre of eye-bar head = 23,620 x 0.875 = 20,670 in. lbs. 
Vertical “ “ post pin plate = 23,620 x 1,875=44,290“ “ 

Inclined “ “ “ = 38,720 x 1.00 = 38,720 “ “ 

The resultant of the last two moments is shown by Fig. 8 
to be: 

R = 37,000 inch-pounds.(9). 

The moment of the greatest vertical component in brace 7 
at the bottom of post 8 is: 

30,980 x 2.6 = 80,550 inch-pounds. . (10). 


Eqs. (6), (7), (9), and (10) show moments far below the 
allowed resisting capacity of a 4 T \ wrought-iron pin, i. e., 
108,000 inch pounds. 

Hence, at C, L, K, and J 5-J inch pins will be used , while at 
D, E, and /, 4^ pins will be taken. 

Before determining the diameters of the pins in the inclined 

end post, it will be necessary to fix 
the sections of that member, and it 
will be convenient to find those of 
the upper chord at the same time. 
As each upper chord panel is a 
beam of considerable span, carry¬ 
ing its own weight, the depth should 
not be small, and it will be taken 
at eighteen (18) inches. The radius 
of gyration of the normal section 
about a horizontal axis through its centre of gravity must 
first be found. 


!c\ 

*r 

_IK*- 

1 

1 



G ^ -H 

00 

7 

1 

'D 


1 

'l 1 


Fig. 9. 














COMPLETE DESIGN OF A RAILWAY BRIDGE. 405 

As the upper chord stress in panels 3 and 4 is about 
322,000 pounds, the area of that panel section will not be far 
from 44 square inches. Fig. 9 represents a trial section of 
that area. 

AB is a 21 x t 9 6 inch cover plate. 

C and D are 3x3 “ 21 lb. angles. 

E “ F “ 5x3 “50 “ 

G “ H “ 18 x i “ side plates. 

The 5-inch legs of E and F are horizontal. The centres 
of gravity of C and D are 0.9 inch from lower surface of AB , 

and those of E and F are 0.9 inch from lower surface of the 

horizontal 5-inch legs. Static moments about a horizontal 
line through the centre of gravity of the section of AB give: 

3x3 angles . . . 2 x 2.1 x 1.2 = 5.04 

5x3 “ . . . 2 x 5.0 x 174 = 174.00 

Side plates ... 18 x 9.3 = 167.4 


Total.346.44 


Hence, 346.44 -f- 44 = 7.9 inches; or the centre of gravity 
g of the entire section is 7.6 inches from the lower surface of 
AB. In such computations some dimensions are taken a 
little full because adjacent surfaces do not have mathemati¬ 
cal contact. 

The elements of the moment of inertia of the section 
about the horizontal axis GH through g take the values : 


Cover plate 
3x3 angles 


11.81 x 7.9 s = 737.06 


2 

X 

2.0 = 

4.0 

4.2 

X 

67 2 = 

188.54 

2 

X 

4-5 = 

9-0 

10.0 

X 

9 - 5 2 = 

902.5 


Side plates 



Moment of inertia 


18 2 4- 12 = 486.0 
18 x 1.4 2 = 35.28 

. . . . 2,362.38 


The moment of inertia of AB about a horizontal axis 
through its own centre of gravity is so small that it has been 
neglected. The 3x3 and 5x3 angles each have a moment 








406 complete design of a railway bridge. 


of inertia of 2 about a horizontal axis through the centre 
of gravity of each respective section. The least radius of 
gyration about a horizontal axis for the entire section then 
becomes: 

V2,362.38 A- 44 — 7-33 inches. 

The panel length is 20.55 ft. Hence 

l+r = 20.55 xi2r 7.33 = 33-7- 

The preceding value in Eqs. (1) gives: 

Flat ends. Pin ends, 

p — 7,620 lbs. per sq. in. / = 7,510 lbs. per sq. in. 

For one pin and one flat end . p — (7,620 + 7,510) - 4 - 2 

= 7,565 lbs. per sq. in. 

The upper chord will be continuous after the bridge is 
erected, but the extremities will be hinged at the upper ends 
of the inclined end posts in the manner shown in Fig. 4 of 
PI. XI. Hence upper chord panel 1 will have one pin end 
and one flat end ; all other panels will be flat end columns. 
The upper chord sections will now be as follows: 

Upper chord 1. 

Required area = 225,424 -4- 7,565 = 30.0 sq. ins. 

1 — 21 x inch cover plate. 9.2 “ “ 

2 — 3 x 3 “ 18 lb. angles 3.6 “ “ 

2 - 5 x 3 “ 30 “ “ 6.0 “ “ 

2 — 18 x | “ side plates. 13.5 “ “ 

Total.32.3 “ “ 

Upper chord 2. 

Required area = 289,238 - 4 - 7,620 = 38.0 sq. ins. 

1 — 21 x inch cover plate . 9.2 “ “ 

2 — 3 x 3 “ 18 lb. angles. 3.6 “ “ 

2 — 5 x 3 “ 36 “ “ 7.2 “ “ 

2 — 18 x 1 “ side plates.. 18.0 “ “ 

38.0 “ “ 


Total 








COMPLETE DESIGN OF A RAILWAY BRIDGE. 


Upper chord 3 and 4. 

Required area = 322,220 -r- 7,620 = 42.3 sq. ins. 

1 — 21 x T \ inch cover plate = 9.2 u “ 

2 — 3x3 “ 18 lb. angles 3.6 “ “ 

2 - 5 x 3 “ 47 “ “ 9.4 “ “ 

2 - 18 x “ side plates. 20.25 “ “ 

Total.42.45 “ “ 

The end post bears on pins at top and bottom ; hence it is 
a pin-ended column. It is about 408 inches long, and it will 
be most convenient to take its depth identical with that of 
the upper chord, or 18 inches. The radius of gyration may 
then be taken, as before, at 7.33 inches. Hence, l a- r — 408 
-f- 7.33 = 55.7. The second formula of Eq. (1) then gives: 

p — 7,070 lbs. per sq. in. 


Inclined end post. 

Required area = 217,750 -7- 7,070 = 

1 — 21 x T \ cover plate_ 

2 — 3 x 3 18 lb. angles .. 
2 - 5 x 3 30 “ 


u 


2 — 18 x | side plates 


30.8 sq. ins. 

9.2 “ « 

3.6 “ “ 

6.0 “ “ 

13.5 “ “ 


Total. 32.3 “ 11 

All these actual areas agree sufficiently near in character 
and amount with the trial section to make re-computations of 
the radius of gyration quite unnecessary. A very little ex¬ 
perience makes such a result possible in all ordinary cases. 
A very close but approximate rule for all box or semi-closed 
sections like those just considered, is to take the radius of 
gyration at four-tenths (0.4) the depth of the side plates. In 
the present case it would make r — 0.4 x 18 = 7.2 inches, while 
the exact value is 7.33 inches. 

In building a section such as these, the angles C and D, 
Fig. 9, should be made as light as possible, in order that the 
cover plate AB may be, to a considerable extent, balanced by 
the heavy angles E, F. In this manner the centre of gravity, 






408 complete design of a railway bridge. 


g, of the section may be brought down sufficiently near to 
the mid depth to give all the space needed inside the chord 
for the eye-bar heads, if the pin axis should be made to pass 
through g, at the same time there is gained the incidental but 
important advantage of an increased moment of inertia. If 
the chord were subject to no bending from its own weight, 
the axis of every pin should pass through the centre of gravity 
of the section. It has been shown in Art. 69 that this flexure 
cannot be satisfactorily neutralized by the direct stress, par¬ 
ticularly if the chord is continuous, as in the present case. It 
is best, therefore, to reduce the bending stresses by making 
the chord depth as great as possible. For these reasons it 
was taken at eighteen (18) inches. If the panels were non- 
continuous the greatest stress per sq. in. in the exterior fibres 
of panels 3 and 4 would be : 


K = 


Md 

I 


88,800 x 10.4 
2,362 


= 390 lbs. 


Now, when it is remembered that the chord is continuous 
it is evident that flexure may be neglected in the sections 
found. This point, however, should always receive careful 
attention. 

In the present case, the axes of pins in the upper chord and 
end posts will he placed eight (8) inches from the line AB , thus 
allowing a small counter moment from the direct stress due 
to a lever arm of 0.4 inch. 

A compression member with the same degree of end con¬ 
straint in all directions ought to have equal capacity for re¬ 
sistance in all directions. If the radius of gyration be taken 
in different directions about g, Fig. 9, for the different sec¬ 
tions as formed, it will be found that this condition is fulfilled. 

Finally, the unsupported width of any plate in compression, 
measured transversely between rivet heads, should not be 
more than about thirty times the thickness. An examination 
of the sections will show that this condition also has been 
fulfilled. 

The details about the pin bearings at the upper and lower 
ends of the inclined end post may now be considered. 




COMPLETE DESIGN OF A RAILWAY BRIDGE. 409 

Fig. 4 of PI. XI., shows the detail at the upper end of the 
inclined end post. The end panel of the upper chord does 
not rest its extremity immediately against the upper end of 
brace 1, but they are separated along the line ab by about 
the distance of -§ inch, and each bears directly against the 
end pin. The diameter of the latter is taken by trial at 5^ 
inches. By the specifications the bearing value of this pin 
against a one-inch plate is 5| x 12,000 = 61,500 pounds. 
Hence for plates T 5 ^, f, 1 and T 9 T inch the bearing values will 
be as follows: 

61,500 x i = 15,375 pounds. 

“ x T 5 g = 19,220 “ 

“ x | = 23,060 “ 

“ x 4 = 30,750 “ 

“ X 16 = 34>596 “ 


The arrangement of thickening plates for upper chord 1 is 
clearly shown by Fig. 4; there is a half-inch plate inside and 
next the fths web and a f jaw plate inside the half-inch thick¬ 
ener. Against the web outside is a T 9 ^ inch thickener. The 
total bearing thickness is then | + i 4- | + = iff inches. 

Hence 61,500 x i|| = 111,500 pounds. The half of the stress 
in upper chord 1 is 112,712 pounds and the two quantities 
are sufficiently near in amount. All rivets about the joint * 
are f inch in diameter. The shearing resistance of one rivet 
at 7,500 pounds per sq. in. is 3,300 pounds, while the bearing 
values against the various plates are: 


3 

¥ 

u 


u 


rivet against T 5 F plate = 2,800 pounds. 

= 3*400 
= 4*500 


u 

u 


u 

<< 


3 

8 

1 

2 


u 




u 


The total bearing pressure against the jaw and thickening 
plates is 23,060 + 30,750 + 34*59^ = 88,406 pounds, and there 
are 21 rivets through those plates, as shown in Fig. 4. Ap¬ 
plying the bearing and shearing values given above to the 
number and distribution of rivets located in that figure, it 
will be seen that there is a little excess of both those 
resistances. 


410 COMPLETE DESIGN OF A RAILWAY BRIDGE. 


The same figure shows the number and distribution of both 
rivets and thickening plates at the upper end of the inclined 
end post. There is a f inch jaw plate outside, then a half¬ 
inch thickener and a T 5 ^ plate between that and the web. 
There is also a quarter-inch thickener inside. The amount of 
bearing thickness is thus the same as for the upper chord. 
An examination of the number and location of the rivets will 
show that there is again a little excess in the bearing and 
shearing resistances. The pitch of rivets in the immediate 
vicinity of the joint is three (3) inches , in all other parts of the 
upper chord and end posts it will be six (6) inches. 

The preceding arrangement is for one side of the chord and 
end post, since both sides, of course, are alike. As Fig. 4 
shows, the pin passes through the jaw plates only. The four 
(4) jaw plates hold the post and upper chord securely to¬ 
gether in case of any derailment or other accident tending to 
knock the end post out of place. They are further reinforced 
by the light -| inch cover plate shown at a. The latter also 
performs an important office in transferring upper lateral 
loads to the end posts. One portion of it, or both, must, of 
course, be riveted in the field. It will be observed that each 
jaw plate has a “ play ” or clearance of i inch, to provide for 
imperfections of workmanship and secure ready erection. 

' The figure shows what rivets must be countersunk, both out¬ 
side and inside. 

The eye-bars of brace 3 lie adjacent to the interiors of the 
upper chord and end post, while those of brace 2 are inside 
of the first. Assuming that the greatest stresses in those 
braces occur together (which is a small error on the side of 
safety), the end pin will be subjected to the bending moments 
shown in Fig. 10. The component moments are as follows: 

167,800 


For brace 3< 


2 


x 1.71 = 143,470 in. lbs. 


u 


u 


45,553 


x 2-9° = 66,053 


(( U 


U 


u 


217,750 X 0.62 — 67,500 “ “ 





COMPLETE DESIGN OF A RAILWAY BRIDGE. 411 

The latter moment arises from the fact that the upper 

chord and end-post bearings have their centres separated by 
0.62 inch. 

In the figure, be is normal to brace 3, and ac is horizontal, 
while ad is normal to 
brace 1. Hence db is the 
resultant moment of 220,- 
000 inch pounds. A pin 
5f inch in diameter will a 
little more than supply 
the required resistance 
with K — 15,000, as Eq. 

(1) of Art. 74 demon¬ 
strates, or as may more 
simply be found by reference to any reliable table of pin mo¬ 
ments. The thickening plates and rivets just found will now 
be a very little excessive, but they will be retained. 

These large rounds frequently vary in standard sizes by 
quarter-inches, and a 54 inch diameter may be turned to 5 f 
with little waste. 

Fig. 7 of PL XI. shows the lower end of the inclined end 
post with the number and location of the f rivets and thick¬ 
ening plates. The operation of designing them is precisely 
similar to those already employed, and they will not now be 
repeated. An examination of the plates and rivets in con¬ 
nection with the preceding values, will show that the shear¬ 
ing and bearing resistances of both the rivets, plates, and 5| 
inch (assumed) pin required by the specifications are secured. 
The line ab is 2 \ inches below the centre of the pin-hole, 
giving about 2 inches of solid metal below the pin. 

Figs. 8, 9 and 10, of PI. XI., show two elevations and a 
plan of the pedestal at the lower extremity of each end post. 
The centre of the pin-hole is taken six (6) inches above the 
bottomf inch plate. The figures show with perfect clearness 
the arrangement of the various parts. The 4 and 19 inch 
spaces give ample clearance for the sides of the end post 
which enter them. 

The vertical component of the end-post stress is 172,820 





412 COMPLETE DESIGN OF A RAILWAY BRIDGE. 


pounds. The total bearing thickness under each half of the 
pin is -| + | + | — if inches. But 64,500 x if ~ 104,610; or 
greater than 172,820 -r- 2 = 86,410. Hence, ample bearing 
surface at 12,000 pounds per square inch is secured. Now 
since the half of the end-post bearing is at the centre of each 
of the 4 inch spaces, it may at first sight appear as if either 
equal bearing areas ought to be found each side of those 
spaces, or as if all ought to be on one side. But it is better 
to mass the metal as much as possible; at the same time the 
weight should be distributed somewhat on the f inch plate. 
The arrangement shown accomplishes these results and gives 
a little excess of bearing area. The 5 by 3 inch angles are 
but 15 inches long, while the f- inch bottom plate is 24 by 36 
inches. 

The bearing thickness at a is 1.25 inches; hence the up¬ 
ward pressure at that surface is 64,500 x 1.25 = 80,625 
pounds. 

The 4 inch space gives about ff inch total clearance for the 
side of the end-post and the eye-bar (if inches thick) of lower 
chord panel 1, or £ inch each for the three clearance spaces 
thus formed. 

The pin moment about the centre of the end-post bearing 
is: 

80,625 x 1.875 inches = 151,171 inch lbs. . . . (11). 

Again, taking moments about the centre of the angle bear¬ 
ing b, there are two moments with horizontal axes but with 
opposite signs formed by the upward pressure at a , and the 
half vertical component in the inclined end post, thus: 

+ 80,625 x 4.81 — + 387,806 in. lbs. 

— 86,413 x 3.00 = — 259,230 “ “ 

Resultant = + 128,576 “ “ 

The stress in the 5 x if inch eye-bar of lower chord 1 has 
the following moment about a vertical axis passing through 
the centre of b: 

65,760 x 1.14 = 74,970 in. lbs. 



COMPLETE DESIGN OF A RAILWAY BRIDGE. 413 

Hence the resultant moment about the centre of b is: 

V(128,57b) 2 + (74, gyof = 150,000 in. lbs. . . . (12). 

As the moment (11) is greater than (12) and far less than 
the resisting capacity of the assumed 5! inch pin, the latter 
diameter will be retained . A smaller pin would give sufficient 
bending resistance, but would necessitate additional metal in 
the thickening plates, and would increase the variety in pins 
and pin-holes. 

It is frequently desirable to hang one pair of eye-bars 
(either braces 2 or 3) outside of chord and end post at the 
upper end of the latter. In such a case the angle flanges at 
by Fig. 4, PI. XI., would be cut away, and more rivets would 
need to be countersunk about the pin-hole on the outside of 
the outer jaw plate. 

In the present instance, however, the pin necessary at the 
upper chord end is but little different from those required by 
braces 3 and 5. Hence, for the sake of uniformity, the pins 
at By Cy Ay A, K and J f will be given a diameter of 5J inches, 
while the others are 4^ inches. 

The pin-plates at the upper and lower end of the intermediate 
posts will now be found. 

It will be assumed that the maximum post stress and the 
greatest panel load occur together. This is not possible, but 
it is difficult to determine the exact maximum load on the 
lower pin-plate, and the assumption involves a safe error. It 
will farther be assumed that the greatest panel load for the 
intermediate posts is the same as the greatest load on brace 
2. This also involves a slight safe error. 

In consequence of these assumptions and the additional 
fact that the smaller part of the load in each lower pin-plate 
is the panel moving load, no addition for impact will be made 
in fixing the thickness of the pin-plates. 

The manner of supporting the ends of the floor-beams is 
clearly shown in Figs. 14, 15 and 16, PI. XI. They are built 
into the posts below the pin. The channels are continued 
30 inches below the centres of the pin-holes, and a 4 by 4 inch 



414 COMPLETE DESIGN OF A RAILWAY BRIDGE. 


36 pound angle is riveted to each as shown at a a Fig. 14* 
The end stiffeners of the floor-beam (Fig, 2, PI. XI.) are 
brought against these latter and riveted fast to them in erec¬ 
tion. The number of rivets required for this connection will be 
found later on. In order to freely admit the end stiffeners 
of the floor-beam, the 10 inch channels of the post will be 
separated 10 inches. 

Vertical Brace 4. 

The top pin plates will carry 99,819 lbs. 

“ bottom “ “ “ “ 99,816 + 45,553 = 145,372 lbs. 

Since 5| x 12,000 = 64,500 pounds, the total thickness of 
bottom pin plates will be 145,372 -4- 64,500 = 2.25 inches ; and 
since the shearing resistance of one J inch rivet is 3,300 
pounds, the total number of rivets in the channel flanges will 
be 145,372 -4- 3,300 = 44. The lower part of Fig. 14, PL XI. 
shows the required arrangement of pin plates and rivets 
There are three -| inch outside pin plates on each side of the 
post. The rivets about the pin-hole on the outside will be 
countersunk in order that the eye-bar of brace 3 may lie 
close against the post. 

The total thickness of pin plates at the top of the post will 
be 99,819 - 4 - 64,500 = i t 9 6- inches, and the total number of 
rivets, 99,819 -4- 3,300 = 30. The upper part of Fig. 14, PI. 
XI. shows the required arrangement of pin plates and rivets. 
As the total number of the latter must be divided by 4, 32 
rivets are used. There is one f inch outside pin plate and 
one T V inch inside plate riveted to the former. The object of 
placing the latter inside is to keep the upper chord as narrow 
as possible. 

Vertical Brace 6. 

The top pin plates will carry 66,193 lbs. 

“ bottom “ “ “ “ 66,193 + 45,553 = 111,746 lbs. 

The total thickness of bottom pin plate will be 111,746 -4- 
64,500= if inches, and that of the upper 66,193-4-50,250 


COMPLETE DESIGN OF A RAILWAY BRIDGE. 415 

= 1 iS' inches, since the upper pin is 4inches in diameter 
and 4 T 3 g- x 12,000 — 50,250 pounds. The total numbers of | 
rivets below and above, respectively, are 111,746 -4- 3,300 = 34 
and 66,193 -4- 3,300 = 20. Fig. 15, PI. XL, shows the required 
pin plates and rivets. At the bottom there is a f inch plate 
next to the channels and a half-inch plate outside. At the 
top there is a | inch plate outside and a T 5 g- inch plate inside, 
as shown. 

Vertical Brace 8. 

Four adjustable ties meet the upper extremity of this post, 
and it has already been shown that each tie adds 4,000 
pounds to the post stress. Hence, the 

top pin plates will carry 34,611 x 16,000 =50,611 lbs. 

bottom “ “ “ 34,611 4- 16,000 + 45,553 = 96,164 “ 

The total thickness of bottom pin plates will be 96,164 -f- 
50,250= I J inches; and that of the top plates 50,611 -4- 
50,250 = I inch. The total numbers of rivets required are 
96,164-4-3,300 = 29 and 50,6114-3,300=16. Fig. 16, PI. 
XI., clearly shows the arrangement of plates and rivets. 
There are more rivets shown at the top than is necessary for 
bearing or shearing alone, for the reason that the notch a 
must be cut out of one channel to let the counterbraces 9 
take hold of the pin inside the post, as there is not room 
enough outside. 


Upper Chord Joints and Thickening Plates. 

It will readily be seen that the pin-hole at the joint point 
between upper chord 1 and 2 is the only one needing a thicken¬ 
ing plate. The greatest tension in an eye-bar of brace 5 is 
119,817 -T- 2 = 59,909 pounds, and the sine of its inclination 
to a vertical line is 0.61. Hence, its horizontal component is 
59,909 x 0.61 = 36,544 pounds. The thickness of the side 
plates of upper 2 is J inch; hence, J x 12,000 x 5§ = 32,250 
pounds. A little over 4,000 pounds, then, is all that need 
be resisted by a thickening plate. This might safely be 


41 6 COMPLETE DESIGN OF A RAILWAY BRIDGE. 


neglected, but the T 5 (T joint plate shown by Fig. 13, PL XI., 
will be extended to cover the pin-hole. 

Precisely the same operation shows that no thickening 
plates are needed at the other pin-holes. 

There will be joints in the upper chord as near as possible 
to, and on the left of the pin-holes at C, D and E of Fig. 1, 
PI. II., and at corresponding points in the other half of the 
truss. 

These joints are formed as shown at Fig. 13, PL XI. At 
C the joint will be 12 inches from the centre of the pin-hole. 
It is formed by riveting top and bottom and side plates to 
the chords, as shown. All the joints are formed precisely 
like this, except that in the other cases the T 5 g- plate between 
the angles extends each side of the outer one, as shown on 
the left only. 

Latticing and Batten Plates. 

The dimensions of latticing and batten plates are matters 
of judgment and experience. Evidently no segment of a 
column between lattice points ought to be less in resistance 
per square unit of section than the column as a whole, but 
experiments are yet lacking to give quantitative results. 
Single latticing with centre lines making angles of 6o° with 
the axis of the member will be used here. 

On the under side of the upper chord and end post the 
lattice bars will be 4 inches by -§, and each end will be held by 

two rivets. Fig. n shows 
this latticing. It will weigh 
about eight (8) pounds per 
lineal foot of member. The 
two battens (one at each 
Fig. 11. end) on the under side of 

the end post and those at the ends of the upper chord (four 
in all) will be 21 inches by 21 inches by f inch. All other 
battens (one on that side of each vertical post opposite to the 
chord joint) will be 21 x 15 x | inches. The bottom plate of 
each joint forms, of course, a batten. 

On the intermediate posts, the latticing will be single and 



















COMPLETE DESIGN OF A RAILWAY BRIDGE. 417 

6o° as before, but the lattice bars will be 2 inches by T 5 ¥ inch. 
This latticing (both sides) will weigh about 9 pounds per 
lineal foot of post. 


Floor-beam Supports. 

The method of suspending the floor-beam from the pin at 
the lower extremity of brace 2 is shown in Fig. 2, PI. XI. 
Two plates riveted to the end stiffeners of the beam take the 
5§ pin with its centre line six inches above the upper flange. 
The greatest moving load carried by the beam end has been 
already found to be 34,600 pounds. One-third of this will be 
added for impact, and as the fixed load is 8,000 pounds, the 
total load to be resisted by the plate hangers becomes : 

- x 34,600 + 8,000 = 54,130 pounds. 

3 

The greatest allowable load per square inch in these hang¬ 
ers is 8,000 pounds; hence the required net area is 54,130 
8,000 = 6.8 square inches. These plates will be taken 12 
inches wide. By deduction of the pin-hole the available 
net width becomes 12 — 5.375 = 6.625 inches. One plate 
12 x Tg- and another 12 x | inch gives the required area. 
Rivets -§ inch in diameter will hold these plates to the end 
stiffeners. The shearing resistance in this case is less than 
the bearing, and the former for one rivet at 7,500 pounds per 
square inch, is 4,500 pounds. Hence the required number 
of rivets is 54,130 -f- 4,500 — 12 rivets. In consequence of 
the deflection of the beam some of the upper ones will be 
subjected to slight tension. Hence 16 rivets are shown. The 
figure shows the number and distribution of rivets and plates. 
The vertical pitch of rivets is 3 inches. 

The manner of attaching the floor-beams to the lower ex¬ 
tremities of the intermediate posts is shown by Fig. 14, PI. 
XI. a and a are 4x4 inch 36 pound angles 27 inches long 
riveted to the inner surfaces of the 10-inch channels, as 
shown. 

The vertical centre lines of the rivet rows in the channels 


27 


4 lS complete design of a railway bridge. 


are coincident with the central lines of the latter, thus insur¬ 
ing an equal division of the floor-beam load between the 
pin-plates. The number and distribution of the -J inch rivets 
in the angles a , a will, of course, be the same as those in the 
plate hangers of Fig. 2, PI. XI. 

Both these methods of supporting floor-beams insure a 
central application to the pin, and the latter insures addi¬ 
tional stiffeners to the floor system and entire structure. 

The two lines of -J inch rivets take less than a square inch 
of section from the two 10 inch 50 pound channels. Hence the 
remaining net section of our nine square inches is more than 
sufficient to carry the total floor-beam load at all points. 

Upper Lateral System. 

A wind pressure of 150 pounds per lineal foot will be taken 
as acting in the horizontal plane of the upper chord. The 
panel wind load will then be 20.55 x 150 = 3,083 pounds. 
Fig. 12 shows a half plan of the upper lateral system. The 
diagonals are ties, and the other members are struts. 

a 

1 

1 

1 



Fig. 12. 


The secant of the inclination of Tj to a horizontal line 
normal to the axis of the bridge is 1.57, and 3,083 x 1.57 = 
4,848 pounds. 

The wind load in the upper chord is a fixed one. The line 
ab is the centre of the span. 

The following stresses may now be written, remembering 
that as the tension diagonals will each be adjustable, 5,000 
pounds must be added for initial stress: 











COMPLE 7 E DESIGN OF A RAILWAY BRIDGE . 419 

P± — — + 5,000 = 5,000 pounds. 

T 3 - 4,848 4 - “ = 9,848 

T 2 = 9,696 + “ = 14,696 “ 

Pi = 14,544 + “ = 19,544 “ 

Pi — 3>°83 + 3,200 = 6,283 pounds 

P 3 = 6,166 + “ rr 9,366 

P 2 = 9,249 + “ = 12,449 

-Pi = 12,332 + “ =15,532 “ 

The greatest allowable stresses in the lateral systems may 
be taken as follows : 

For tension .... 14,000 pounds per sq. in. 

“ compression . . - 9 - ^°^ - « « (43). 

I + - ~b 

30,000 P 

The latter formula is for flat-end members of angle iron, as 
those will be used for lateral compression members. It will 
be observed that it gives less value than the formula (1) for 
columns of the box type, like those used in the trusses. 
Under these stresses the tension members become: 


Ti .I — i-J o. 

T 3 .1 — I “8 o. 

T2 .1 — i£ o. 

Pi .1 - if o. 


It is not advisable to have any tension member less in sec¬ 
tional area than 1 square inch. Hence, T 3 and T 4 are a little 
larger than the stresses require. 

All the struts except P x will be formed of 3 — 3 x 3 inch 
angles. A section of this strut is shown at Fig. 11. PI. XI. 
The two angles c lie on the upper chord and are riveted to 
it, as shown. These two angles are designed to carry all the 
stress of the strut. The only office of the angle a is to keep 
the strut stiff in a vertical plane; it takes hold of the lower 
flange of the chord with two rivets, as shown at c , Fig. 4, PI. 








420 COMPLETE DESIGN OF A RAILWAY BRIDGE. 


XI. The strut is thus of the same depth as the chord, and 
takes hold of both those members in such a manner as to 
give them great rigidity. At each end of the strut there is a 
20 x 12 inch plate, and there is a set of single 6o° lacing, 
2 x T V inch. The radius of gyration of the section of the 
two top angles about a vertical line midway between the two, 
is 1.3 inches. The length of the strut is about 192 inches. 
Hence Eq. (13) gives 5,300 pounds as the greatest allowable 
stress per square inch. As P 2 requires only the lightest 
angle that ought to be used, all these struts will be made 
alike. 




As the detail for these struts would give some trouble at 
the end of the upper chord, P x will be a single 6 " x 4’' angle 
with the 6" leg horizontal, as shown at a, Fig. 4, PL XI. ' 

The tie 7 j is attached to the upper chord by the detail 
shown at b, Fig. 11, PI. XI. A piece of 6 " by 4", 60 pound 
angle, 12 inches long, with the 6 inch leg lying on the cover 
plate of the chord, carries two pieces of 3" by 3", 21 pound 
angles about 5J- inches long, and with edges parallel to the 
axis of T v One end of each of the latter angles rests squarely 
against the vertical 4-inch leg of the 6" by 4" angle. Six three- 
quarter inch rivets are then passed through the angles in 
the manner shown. Each such rivet will resist about 4,000 
pounds in single shear. The tie 7 \ passes through the 4-inch 
leg of the heavy angle (between the 3-inch angles), and carries 
a nut at its end, which gives the requisite adjustment. 

T 2 and all the other lateral ties are held by the same detail, 
except that 4 rivets only are needed, as shown at b'. 


Transverse Bracing. 


A skeleton sketch of the intermediate transverse bracing 
for the vertical plane of any two opposite posts is shown in 
Fig. 16, PI. XII. If the upper lateral system is designed to 


COMPLETE DESIGN OF A RAILWAY BRIDGE . 421 


carry the whole wind load to the ends of the upper chord, as 
has been supposed, the duty of the intermediate transverse 
bracing is entirely indeterminate. As the sections must be 
determined in some manner, however, the method of Art. 
82, Fig. 1, will be applied. 

The slight analytical superabundance of stability thus 
secured is no more than is required by a rapidly moving 
load. 

F' of Art. 82 will here be taken as 3,083 pounds and P= o. 
Also, b — 17 ft. and a = 6 ft. It will here be assumed that 
all the wind load is applied in the windward truss. This is 
the usual assumption in practice, although the conditions 
taken in Art. 82 are exactly true. Eq. (1) of that Art. then 
gives: 

w — (3,083 x 27) -r- 17 = 4,932 pounds. 

As the tangent of the inclination of T to a vertical line is 
2.833 ar, d the secant, 3.0, the stresses are: 

P=4,932 x 3 4- 5,000= 19,796 pounds.if o. 

P= 4,932 X 2,833 + 5,000= 18,974 “ 2-3" x 3"24 lb. angles. 

The sizes are based on the same allowed working stresses 
as for the upper laterals. The strut Pis shown at Fig. 19, 
PI. XII. The two angles are held if inches apart by separa¬ 
tors, and present a horizontal upper surface. The ends are 
secured to a batten plate in proper position on the post and 
in the manner shown. The separation of the angles permits 
the tie T to pass between them and through the batten plate 
and take a nut inside the post. The washer# is formed from 
a piece of 3 by 2 inch angle, one-half inch thick, with the 2 
inch leg sheared off until the proper angle is formed. The 3 
by 2 inch angle b forms a check to keep the angle washer a 
in place. 

The length of the strut Pis 192 inches, while the radius of 
gyration of a 3 by 3 inch angle about an axis through its 
centre of gravity and parallel to one leg is 0.92 inch. Hence 
the allowable stress per sq. in. by Eq. (13), is 4,000 pounds. 


422 COMPLETE DESIGN OF A RAILWAY BRIDGE. 

Both ends of T are held by precisely the same detail. At 
the upper end of the post, however, the pin-plates take the 
place of the battens at the intermediate points. The rivets 
securing the batten plate to the post are seen to give an ex¬ 
cess of resistance. 

Portal Bracing. 

Fig. 17, PI. XII., shows a skeleton sketch of the portal 
bracing. The sketch is taken in the plane of the portal. The 
strut P is placed 7 ft. 6 in. from the top of the post, while the 
length of the latter is 33.8 ft. 

The computations are made precisely as in connection with 
Fig. 16, PL XII. The force acting at the upper extremity of 
the end post is 12,332 pounds. The tangent of the angle be¬ 
tween T and the end post is 2.27, while the secant is 2.48. 
Hence, if w — 12,332 x 33.8 -f- 17 = 24,664, then : 

T — 24,664 x 2.48 = 61,170 pounds.. 1 — 6" x 4" 60 lb. angle. 
P — 24,664 x 2.27 = 55,990 “ ..1— 6 " x 6 " 75 “ “ 

As shown by the preceding results, this bracing is com¬ 
posed entirely of angles. This is done in order to secure the 
utmost stiffness or rigidity in the portal. 

Fig. 12, PI. XI., shows the method of securing the ends of 
the members T and P. The upper extremity of T is shown 
with the six-inch leg of the angle lying on the end post at a. 
In order that the proper number of three-quarter inch rivets 
may be brought into play, a -§ inch plate lies underneath the 
angle, as shown. The method of securing the lower end of 
T, and each end of Pis clearly shown at b. Three-quarter 
inch rivets are used for all these connections. At the inter¬ 
section of the two Ts, one is cut and a firm joint is made by 
a centre plate a half-inch thick, aided by angle lugs. 

The length of the strut P is about 192 inches, and its radius 
of gyration, 1.9 inches. Hence Eq. (13) gives the working 
stress at 7,000 pounds per square inch. The tension allowed 
in this angle bracing is 12,000 pounds per square inch. 

An ornamental wrought-iron bracket may be placed in the 
angle between P and the end post. 


COMPLETE DESIGN OF A RAILWAY BRIDGE. 423 
Lower Lateral Bracing. 

The lower lateral bracing is shown in skeleton plan by Fig. 
6, PI. XII. It is designed to resist a uniform fixed wind load 
of 150 pounds per lineal foot in addition to a uniform moving 
wind load of 300 pounds per lineal foot. The fixed panel 
load, therefore, will be 20.55 ><150= 3,083 pounds, and the 
moving panel load, 20.55 x 3 °° — 6,166 pounds. The secant 
of the angle between T x and P 1 is 1.57. The line ab is the 
centre line of the span. Remembering that there are nine 
panels in the lower lateral system, and that the greatest al¬ 
lowable tension is 14,000 pounds, the following stresses and 
sizes may at once be written from the preceding data: 


7 i stress .58,164 + 5,000 = 63,164 lbs. 1 - 2"} round. 

T 2 “ .44,700+ “ =49,700 “ 1-2"! “ 

T s “ . 32 , 313 + “ = 37»3 1 3 “ i-i| “ 

“ 2I,COO+ “ = 26,000 “ I — I | “ 

T 5 “ .10,770+ “ = 15,770 “ i-ii “ 


The initial stress is included in the total by the addition of 
5,000 pounds, as has been done before. 

The floor-beams form the struts in the lower lateral system, 
except in the cases of the end struts P lf hence, only the latter 
need be provided. The friction on the wall-plate will evi¬ 
dently relieve P 1 of some of its stress, but it is uncertain to 
what extent; hence, initial tension only will be neglected. 
Consequently: 

P x stress .41,620 lbs .1 — 6" x 4" 50 lb. ajigle. 

The 6 inch leg of this angle is horizontal, and it is riveted 
to the top flanges of the stringers where it crosses the latter . By 
this means the stringer ends are held rigidly in position, and 
the general stiffness is increased. 

The rods of the lower lateral system will necessarily pass 
through the webs of the stringers, and will be secured to the 
webs of the floor-beams (as closely as possible to their upper 
flanges) by precisely the detail shown in Fig. 19, PI. XII. 









424 COMPLETE DESIGN OF A RAILWAY BRIDGE. 

The web-plate now takes the place of the batten in that 
figure. 

The lower lateral T x takes hold of the pedestal by the clevis 
and 3 inch pin bolt shown in Fig. io, PI. XI. 

A g" x 7'' x plate is riveted above and below the } inch 
base plate of the pedestal in order to give the proper bearing 
area. The method of securing the end of P 1 to the pedestal 
is shown by the same figure with perfect clearness. 

Expansion Rollers . 

A set of expansion rollers under one end of each truss must 
be provided. A diameter of 2 \\ inches will be assumed. 

According to Appendix II., the resistance per lineal inch 
of a roller is: 

i«\Nm =5v? <,or e= £> 

Since all metal is wrought-iron E — E'. The greatest al¬ 
lowable intensity of pressure on the roller will be taken at 
12,coo pounds per square inch, or w — 12,000. Also R — 1.47 
and E — 26,000,000. Hence the allowable load per lineal 
inch, by the above formula, is 1,050 pounds. The maximum 
vertical component of the end-post stress is 172,820 pounds. 
Hence the number of lineal inches of roller bearing required 
is 172,820 -r- 1,050 = 164 inches. The set of rollers shown by 
Fig. 11, PI. XII., gives very closely the required amount. 
The clear space between each adjacent pair of rollers is § inch. 
The ends of each roller are turned down to -J inch and pass 
through a 2l x | inch wrought-iron strap on the outside of 
which each roller end takes a nut. The rollers are thus held 
rigidly in their proper relative positions. A I x { inch collar 
is turned down at the centre of each roller to take the 1 x T 8 ^- 
inch shoulder which is shown in the wall'plate, and by means 
of which all lateral motion of the rollers is prevented. 

Wall Plates. 

4 

The mean pressure per square inch on the total surface of 





COMPLETE DESIGN OF A RAILWAY BRIDGE . 425 

the wall plate should not exceed 200 pounds. The total area 
of wall-plate surface shown in Fig. 12, PI. XII., is 30 x 30 = 
900; hence the total allowable weight is 900 x 200 = 180,000 
pounds. The maximum total vertical pressure of 172,820 
pounds is thus provided for. 

The plan of the wall plate at the roller end is shown in the 
figure; the upper elevation also belongs to it. At the fixed 
end either the pedestal or the masonry must be sufficiently 
high to fill the roller space. Both those alterations, however, 
are unadvisable for obvious reasons. It is better to fill the 
roller space with the wall plate. The lower elevation of Fig. 
12, PI. XII., shows the arrangement to be adopted. The 
same -J inch thick wall plate is to be taken, 4 — 3J x 3^ x f 
inch angles then run in the direction of the rollers across the 
entire plate. At right angles to these 4 lines of 3 x 3 x J 
inch angles are placed. The latter are cut to fit in between 
the former and will, of course, require filling strips under¬ 
neath. The top of this gridiron arrangement is then planed 
off until the proper height of wall plate is reached. The 
rectangular spaces thus formed are then filled with Portland 
cement rammed hard and flush with the planed upper sur¬ 
faces of the angles. A solid wall plate is thus formed with 
the interior surfaces completely protected against corrosion. 

At diagonally opposite corners are seen the holes for the 
if inch anchor bolts. 

Wind Pressure on Chords and End Posts. 

The effect of the wind load on both upper and lower chords 
has been shown in detail in Art. 81 ; the principles there 
established remain to be applied here. The chord stresses in 
the lateral trusses are the same in kind as those produced by 
the vertical loading in one lower chord and one upper chord. 
Just to what extent these wind stresses may be allowed to 
exist without necessitating any increased chord section is a 
matter of experience only; but as the greatest wind stresses 
and those due to the vertical loading so rarely combine in 
most localities that with the working stresses specified in this . 


4^6 COMPLETE DESIGN OF A RAILWAY BRIDGE. 


case the wind load may be allowed to reach \ths the value 
of the greatest vertical loading without requiring any increase 
in chord section , in all localities not ordinarily subject to 
cyclones or tornadoes. 

In the present instance the total fixed and moving vertical 
load is equivalent to about 2,040 pounds per lineal foot of 
each truss. The total wind load in the lower chord is 450 
pounds per lineal foot, and its depth of truss is only 17 
feet. If reduced to the same truss depth as that for the 


27 

vertical load, i. e., 2 7 feet, it would be 450 x — = 720 pounds. 


Again, the overturning effect of the wind on the train (dis¬ 
cussed at the clos'e of Art. 81) throws on the leeward truss 

the additional weight of = 140 pounds per lineal foot. 

It is assumed that the centre of wind pressure on the train is 
8 feet above the end supports of the floor-beam. The total 
wind effect on the loading of the leeward truss is then 
720 + 140 == 860 pounds per lineal foot, or a little in ex¬ 
cess of four-tenths the vertical loading. As three-eighths 
the vertical loading is 765 pounds per lineal foot, the chord 
sections should be increased for 95 pounds per lineal foot. 
As the increase in area, however, would be but one-sixth 
of an inch for one chord, and as the transverse bracing 
will slightly relieve the leeward truss, no change will be 
made. 

If the lower chord needs no revision, the upper need not 
be considered. 

The total pressure of wind against the upper extremities of 
the end posts, and, hence, against their lower extremities 
also, has already been seen to be 12,332 pounds. If this is 
assumed to be equally divided between the end-post feet, 
each of the latter will carry 6,166 pounds. Each end of 
P (in the portal, Fig. 17, PI. XII.) is 26.3 feet from the end- 
post foot. Hence at the former point the end post suffers 
the bending moment 6,166 x 12 x 26.3 = 1,945,990 in. lbs. 
The moment of inertia of the end-post section about the 
neutral axis normal to the cover plate is 1,886, and since the 



COMPLETE DESIGN OF A RAILWAY BRIDGE. 427 


half total width of the chord is 12.5 inches, the stress per 
square inch in the extreme fibres of the 5x3 angles is 


K' = 1,945,990 x 12.5 -r- 1,886 = 12,904 pounds per sq. in. 


The direct post stress will be about 7,200 more, or a total 
of 20,104 pounds per sq. in ., whereas 15,000 should not be ex¬ 
ceeded. 

If K — limit of compressive bending stress per square inch, 
which must not be exceeded ; M — bending moment in inch 
pounds ; I the moment of inertia of the total post section 
about a neutral axis normal to the cover-plate ; 2 d — total 
width of end post; P — total direct stress of compression, 
due to vertical loading and overturning action of the wind 
against the train (t of Art. 81 is its panel value), and A = 
total area of section ; then the sectional area must be increased 
until the following equation holds true: 


AT = 


Md P 

~r + a 



It has been seen above that the wind effect in the leeward 
truss is 140 lbs. per lin. ft., or 20.55 x l 4° = 2 fi77 lbs. per 
panel. Hence, 4 x 2,877 x l - 2 & = I 4 > 5 00 lbs. is that part 
of P due to the wind ; or: 

P= 14,500 + 217,750 = 232,250 lbs. 

Also, 2d = 25 ; or d — 12.5 inches; 

And, M = 1,945, 990 in. lbs. 

If 2 — 3" x 3" 25 lb. angles are riveted on the outside of 
each side plate of each post in the manner shown in Fig. 13, 
the centres of gravity of those angles will be 8.3 inches from 
the neutral axis about which / is taken, and the moment of 
inertia of each angle section about a parallel axis through its 
own centre of gravity is 2.25; hence: 

/ = 1,886 + 4 x 2.5 x (8.3) 2 + 4 x 2.25 = 2,585. 


Finally: 

A = 32.3 + 4 x 2.5 = 42.3 sq. ins. 




428 COMPLETE DESIGN OF A RAILWAY BRIDGE. 


These quantities placed in Eq. (14) give: 

K — 9,412 + 5,500 = 14,912 lbs. per sq. in . 

which shows that the desired section is obtained. 

Fig. 13 shows an elevation of parts of the end post, which 
is supposed to be intersected by the portal strut at ab. cc ', and 
dd' are the 3" x 3" 25 lb. angles, c is 9 ft. below ab and c' 
2 ft. 9 in. above it. dd' is half the length of cc'. Below c 
and above c', the preceding figures show that no increase of 
section is needed. 

If much increase is needed, unequal legged angles with 




the larger legs normal to the side plates can be most advan¬ 
tageously used. 

It will ordinarily be sufficiently accurate to increase the 

K’ 

section in the ratio of -=• 

K. 

These computations show what an important factor the 
wind load may be in a country subject to tornadoes and cy¬ 
clones. In such exposed localities the chord-wind stresses 
should not be allowed to exceed 25 per cent, of those due to 
the vertical loading without providing correspondingly in¬ 
creased sections. 

The wind effect on the stringers mentioned in Art. 81 is 
such a small per centage of the vertical load that it need not 
be considered. 

Conclusion . 

It is not necessary here to produce in detail the complete 
list of weights of all the parts, although this must invariably 













































COMPLE TE DESIGN OF A RAIL WA Y BRIDGE . 429 


be done in practice. If the estimated weight comes out 
greater than the assumed, a revision of the computations 
must be made with a sufficiently increased fixed weight to 
exceed, at least by a little, the estimated weight. If, on the 
other hand, the estimated weight is considerably less than 
the assumed, the latter may be reduced in a recomputatiom 
in order to reach a proper degree of economy. 






/ 





APPENDIX I. 


THE THEOREM OF THREE MOMENTS. 

Art. i.—T he object of this theorem is the determination 
of the relation existing between the bending moments which 
are found in any continuous beam at any three adjacent 
points of support. In the most general case to which the 
theorem applies, the section of the beam is supposed to be 
variable, the points of support are not supposed to be in the 
same level, and at any point, or all points, of support there 
may be constraint applied to the beam, external to the load 
which it is to carry; or, what is equivalent to the last condi¬ 
tion, the beam may not be straight at any point of support 
before flexure takes place. 

Before establishing the theorem itself, some preliminary 
matters must receive attention. 

In Fig. i, let ABC represent the centre line of any bent 
beam; AF, a vertical line through A ; CF 1 a horizontal line 



through C, while A is the section of the beam at which the 
deflection (vertical or horizontal) in reference to C, the bend¬ 
ing moment, the shearing stress, etc., are to be determined. 

43i 





432 


APPENDIX I. 


As shown in the figure, let ;tr be the horizontal co-ordinate 
measured from A, and y the vertical one measured from the 
same point; then let z be the horizontal distance from the 
same point to the point of application of any external ver¬ 
tical force P. To complete the notation, let D be the deflec¬ 
tion desired ; M lt the moment of the external forces about A ; 
S, the shear at A ; P', the strain (extension or compression) 
per unit of length of a fibre parallel to the neutral surface 
and situated at a normal distance of unity from it; /, the 
general expression of the moment of inertia of a normal cross- 
section of the beam, taken in reference to the neutral axis of 
that section ; E the coefficient of elasticity for the material 
of the beam; and M the moment of the external forces for 
any section, as B. 

Again, let A be an indefinitely small portion of any normal 
cross-section of the beam, and let y be an ordinate normal 
to the neutral axis of the same section. By the “ common 
theory ” of flexure, the intensity of stress at the distance y 
from the neutral surface is ( y PE ). Consequently the stress 
developed in the portion A, of the section, is EP'y'A , and the 
resisting moment of that stress is EP’y" l A. 

The resisting moment of the whole section will therefore 
be found by taking the sum of all such moments for its 
whole area. 

Hence: 

M = EP'Zy’^A = EP'L 


Hence, also, 


, _ M 

El ‘ 


If n represents an indefinitely short portion of the neutral 
surface, the strain for such a length of fibre at unit’s distance 
from that surface will be nP'. 

If the beam were originally straight and horizontal, n would 
be equal to dx. 

P' being supposed small, the effect of the strain 71P ' at any 
section, B } is to cause the end K, of the tangent BK, to move 
vertically through the distance nP'x. 



THEOREM OF THREE MOMENTS. 


433 

If BK and BR (taken equal) are the positions of the tan¬ 
gents before and after flexure, nP x will be the vertical dis¬ 
tance between K and R. 

By precisely the same kinematical principle, the expression 
nP'y will be the horizontal movement of A in reference to B. 

Let 2 ?nP'x and 'EnP'y represent summations extending 
from A to C, then will those expressions be the vertical and 
horizontal deflections, respectively, of A in reference to C. 
It is evident that these operations are perfectly general, and 
that x and y may be taken in any direction whatever. 

The following general, but, strictly, approximate equations, 
relating to the subject of flexure, may now be written : 


5 - 2 P . . 


.... (i). 

Mi - 2 Pz . . 


.... (2). 

• 

• 

ii 


.... ( 3 ). 

M 

2 nP' = 2 n * 
hi 


.... ( 4 ). 

D = SnP'x = 2 

nMx 

El 

1 

. . . . (5). 

D h — 2 nP'y — 2 

nMy 

El 

. . . . (6). 


D h represents horizontal deflection. 

Art. 2.— Some elementary but general considerations in 
reference to that portion of a continuous beam included be¬ 
tween two adjacent points of support must next be noticed. 

If a beam is simply supported at each end, the reactions 
are found by dividing the applied loads according to the 
simple principle of the lever. If, however, either or both 
ends are not simply supported, the reaction, in general, is 
greater at one end and less at the other, than would be found 










434 


APPENDIX /. 


by the law of the lever; a portion of the reaction at one end 
is, as it were, transferred to the other. This transference can 
only be accomplished by the application of a couple to the 
beam, the forces of the couple being applied at the two adja¬ 
cent points of support ; the span, consequently, will be the 
lever arm of the couple. The existence of equilibrium re¬ 
quires the application to the beam of an equal and opposite 
couple. It is only necessary, however, to consider, in connec¬ 
tion with the span AB , the one shown in Fig. 2. Further, 
from what has immediately preceded, it appears that the 
force of this couple is equal to the difference between the 
actual reaction at either point of support and that found by 
the law of the lever. The bending caused by this couple will 
evidently be of an opposite kind to that existing in a beam 
simply supported at each end. 

These results are represented graphically in Fig. 2. A and 
B are points of support, and AB is the beam; AR and BR' 
are the reactions according to the law of the lever; RF = 



RF is the force of the applied couple; consequently AF = 
AR + RF and BF— BR' — ( R'F — RF) are the reactions 
after the couple is applied. As is well known, lines parallel 
to CK, drawn in the triangle A CB } represent the bending mo¬ 
ments at the various sections of the beam, when the reac¬ 
tions are AR and BR'. Finally, vertical lines parallel to AG> 









THEOREM OF THREE MOMENTS. 


435 


in the triangle QHG , will represent the bending moments 
caused by the force R'F. 

In the general case there may also be applied to the beam 
two equal and opposite couples, having axes passing through 
A and B respectively. The effect of such couples will be 
nothing so far as the reactions are concerned, but they will 
cause uniform bending between A and B. This uniform or 
constant moment may be represented by vertical lines drawn 
parallel to AH or LN (equal to each other) between the lines 
AB and HQ. dhe resultant moments to which the various 
sections of the beam are subjected will then be represented 
by the algebraic sum of the three vertical ordinates included 
between the lines ACB and GQ. Let that resultant be 
called M. 

Let the moment GA be called M a , and the moment BQ = 
LN = HA,M b . Also designate the moment caused by the 
load P, shown by lines parallel to CK in ACB, by M x . Then 
let x be any horizontal distance measured from A towards B ; 
/ the horizontal distance AB; and z the distance of the point 
of application, K , of the force P from A. With this notation 
there can be at once written: 

M=M a (PNj + M b Q + M, . . . (7). 

Eq. (7) is simply the general form of Eq. (2). 

It is to be noticed that Fig. 2 does not show all the mo¬ 
ments M a , M b and to be of the same sign, but, for con¬ 
venience, they are so written in Eq. (7). 

Art. 3.—The formula which represents the theorem of 
three moments can now be written without difficulty. The 
method to be followed involves the improvements added by 
Prof. H. T. Eddy, and is the same as that given by him in the 
“American Journal of Mathematics,” Vol. I., No. 1. 

Fig. 3 shows a portion of a continuous beam, including 
two spans and three points of support. The deflections will 
be supposed measured from the horizontal line NQ. The 
spans are represented by l a and l c ; the vertical distances of 



43 6 


APPENDIX /. 


NQ from the points of support by c a , c b and c c ; the moments 
at the same points by M a} M b and M n while the letters 5 and 
R represent shears and reactions respectively. 



Fig. 3. 


In order to make the case general, it will be supposed that 
the beam is curved in a vertical plane, and has an elbow at b y 
before flexure, and that, at that point of support, the tangent 
of its inclination to a horizontal line, toward the span l a is t, 
while t' represents the tangent on the other side of the same 
point of support; also let d and d' be the vertical distances, 
before bending takes place, of the points # and c, respectively, 
below the tangents at the point b. 

A portion of the difference between c a and c b is due to the 
original inclination, whose tangent is /, and the original lack 
of straightness, and is not caused by the bending; that por¬ 
tion which is due to the bending, however, is, remembering 
Eq. (5): 


D — c a Cb 


, , , Mxn 

l a t-d=2 b -gj- 


By the aid of Eq. (7) this equation may be written: 





In this equation, it is to be remembered, both x and z (in¬ 
volved in M i) are measured from support a toward support 
b. Now let a similar equation be written for the span / c , in 














THEOREM OF THREE MOMENTS. 


437 

which the variables x and z will be measured from c toward 
b. There will then result: 


E(c c — c b — l c f — d) 




When the general sign of summation is displaced by the 
integral sign, n becomes the differential of the axis of the 
beam, or ds. But ds may be represented by udx, n being such 
a function of x as becomes unity if the axis of the beam is 
originally straight and parallel to the axis of x. The Eqs. 
(8) and (9) may then be reduced to simpler forms by the fol¬ 
lowing methods: 

In Eq. (8) put 



Also, 


^r a u(l a -x)dx = i ^f±f\l a -x)dx = 

4 J b 4 «/ & 

4 %a ^4 4 


• • 


In the same manner: 

a X*n i P a ux*dx x a ’ p (l nxdx 


a x*n _ I_ P 

6 4 J b 


I 


l 


f 

a*s b 


1 


• ( I2 )- 


• • 


• (n)- 


Xa_ n uxcix _ r a uxdx _ _ 

4 «/ b * 4 ts b 


Also, 


• (14)- 



















433 


APPENDIX I 


And, 


7 ' V ' /"* ® 

a \ n -J uxdx 


*' v ’ nm xdx = ENhHl 


^ a 


4 

J 


( 15 )- 


Again, in the same manner: 


-v’' d XII _ • -^71 -n /r a 

2 b — j — = z la u la 2 M t x A x 


(16). 


Using Eqs. (io) to (16), Eq. (8) may be written: 

E (c c Cfy l a t d ) — — (M a u a z a x a -j- AT b u a z a x a ) 4- 

Ula ha 2 1 Mx A X . . . ( 17 ). 

Proceeding in precisely the same manner with the span l & 
Eq. (9) becomes: 

E (c c — c b — l c f — d’) = ~{M c u c i c x c + M b u c 'i c 'x c ') + 

u lc i Xc 2 ° b M x x A x . . . (18). 


The quantities x a and x c are to be determined by applying 
Eq. (10) to the span indicated by the subscript; while u ay i af 
u c and i c are to be determined by using Eqs. (11) and (12) in 
the same way. Similar observations apply to zz a ', i a ', x n ', u c ', 
ic and *c\ taken in connection with Eqs. (13), (14) and (15). 

If I is not a continuous function of x, the various integra¬ 
tions of Eqs. (10), (11), (13), and (14) must give place to sum¬ 
mations (2) taken between the proper limits. 

Dividing Eqs. (17) and (18) by l a and l c , respectively, and 
adding the results: 



+ 


Cb 


L 


~ d d'\ 

“4 “ 47 “ 


M,a l,a 


L 


2 h M x xAx + 


a 


^~^. c b M v xAx + l(M a u a i a x a +M„u c ;iaX a ' + 

LC 

M c u c icX c + M b u c i c 'x c ') .... (19). 


in which T= t + t'. 










THEOREM OF THREE MOMENTS. 


439 


Eq. (19) is the most general form of the theorem of three 
moments if E, the coefficient of elasticity, is a constant quan¬ 
tity. Indeed, that equation expresses, as it stands, the “ the¬ 
orem ” for a variable coefficient of elasticity if ( ie ) be written 
instead of i ; e representing a quantity determined in a man¬ 
ner exactly similar to that used in connection with the quan¬ 
tity i. 

In the ordinary case of an engineer’s experience, T — o, 
d — d'= o, I — constant, u — u a — u c — etc .= c — secant of the 
inclination for which t — — t'is the tangent; consequently 



From Eq. (10), 


2 l c 

x ° 6 1 


% a — 


2 1 , 


a 


From Eq. (13), 

^ a 

Xa — 6 > 



The summation 'EM^xAx can be readily made by referring 
to Fig. 2. 

The moment represented by CK in that figure is, 


P 



consequently the moment 
to P, is, 



at any point between A and K, due 


• ^ • - =p[ l —f)x. 


Between K and B , 

M{ = fff) • CK=P- l {l-x). 


Using these quantities for the span l a : 










440 


APPENDIX I. 


21 xAx =fl M x xdx + f'“M{xdx = \P{Q - s?)e. 

For the span l c , the subscript a is to be changed to c . 

Introducing all these quantities, Eq. (19) becomes, after 
providing for any number of weights, P: 

6£7 + <MlN\ = ^ + 2Ah (4 + 4) + MJc + 

C \ l a "c ' 

~2P(Q-P)z+ ~ i'P(/ c 2 - P ) z . . . . (20). 

Eq. (20), with c’ equal to unity, is the form in which the 
theorem of three moments is usually given ; with c equal to 
unity or not, it applies only to a beam which is straight before 
flexure, since T = t -f t' — o — d — d ] . 

If such a beam rests on the supports a, b, and c , before 

bending takes place, ° a ° b = — C ° and the first mem- 

t a t c 

ber of Eq. (20) becomes zero. 

If, in the general case to which Eq. (19) applies, the deflec¬ 
tions c a , c b , and e c belong to the beam in a position of no 
bending, the first member of that equation disappears, since 
it is the sum of the deflections due to bending only , for the 
spans 4 and l c , divided by those spans, and .each of those 
quantities is zero by the equation immediately preceding, Eq. 
(8). Also, if the beam or truss belonging to each span is 
straight between the points of support ( such points being sup - 
i)osed in the same level or not), u a — u a — u \a — constant, and 
u c — u c ' = u lc = another constant. If, finally, / be again taken 
as constant, x a and x c , as well as ^M x xAx, will have the values 
found above. 

From these considerations it at once follows that the 
second member of Eq. (20), put equal to zero, expresses the 
theorem of three moments for a beam or truss straight be¬ 
tween points of support, when those points are not in the 
same level, but when they belong to a configuration of no 
bending in the beam. Such an equation, however, does not 
belong to a beam not straight between points of support. 







THEOREM OF THREE MOMENTS. 


441 


The shear at either end of any span, as l ay is next to be 
found, and it can be at once written by referring to the ob¬ 
servations made in connection with Fig. 2. It was there seen 
that the reaction found by the simple law of the lever is to 
be increased or decreased for the continuous beam, by an 
amount found by dividing the difference of the moments at 
the extremities of any span by the span itself. Referring, 
therefore, to Fig. 3, for the shears S, there may at once be 
written : 


^ p {a _ 5 _ M a M b 


v a 


v a 


s„'= - M * 


£, „ z M. — M. 


S b = 2P~ + 


s;=sprs- 


4 — Z M r — M h 



. (22). 

• (23)- 

• (24). 


The negative sign is put before the fraction 


M a -M b 

la 


in 


Eq. (21), because in Fig. 2 the moments M a and M b are rep¬ 
resented opposite in sign to that caused by P, while in Eq. 
(7) the three moments are given the same sign, as has already 
been noticed. 

Eqs. (21) to (24) are so written as to make an upward re¬ 
action positive, and they may, perhaps, be more simply found 
by taking moments about either end of a span. For example, 
taking moments about the right end of l a \ 


S a l a -2P(l a -z) + M a = M b . 


From this, Eq. (21) at once results. Again, moments about , 
the left end of the same span give : 

S b la — ^ Pz + M b — M a . 









442 


APPENDIX I. 



This equation gives Eq. (22), and the same process will 
give the others. 

If the loading over the different spans is of uniform inten¬ 
sity, then, in general, P — wdz ; w being the intensity. Con¬ 
sequently : 

2 P(l 2 — z 2 ) z = f w (/ 2 — z 1 ) zdz — w —. 

J 0 4 


1 a 

In all equations, therefore, for -—2 P(l a 2 — z 2 ) z there is to 

l 3 I c 

be placed the term w a — ; and for -~ 2 P(l 2 — z 2 ) z, the term 

4 4 

4 3 

w c — . The letters a and c mean, of course, that reference is 
4 

made to the spans l a and l c . 

From Fig. 3, there may at once be written: 


R — S a ' + S a 


R' = 5 ; + 5 , 
R" = s' + s; 


ctc.= etc. + etc. 


(25) . 

(26) . 
■ ( 2 7h 





APPENDIX II. 


THE RESISTANCE OF SOLID METALLIC ROLLERS. 

An approximate expression for the resistance of a rollei 
may easily be written, and although the approximation may 
be considered a loose one, it furnishes an excellent basis for 
an accurate empirical formula. 

The following investigation contains the improvements by 
Prof. J. B.. Johnson and Prof. H. T. Eddy on the method 
originally given by the author. 

The roller will be assumed 
to be composed of indefi¬ 
nitely thin vertical slices par¬ 
allel to its axis. It will also 
be assumed that the layers 
or slices act independently 
of each other. 

Let E' be the coefficient 
of elasticity of the metal over 
the roller. 

Let E be the coefficient 
of elasticity of the metal 
of the roller. 

Let R be the radius of the 
roller and R' the thickness of the metal above it. 

Let w = intensity of pressure at A . 

“ p — “ “ any other point. 

u P = total weight which the roller sustains per unit of 
length. 

“ x be measured horizontally from A as the origin. 

“ d = AC. 

4 ‘ e — DC. 








444 


APPENDIX II. 


From Fig. i: 


AB = A’B' = p ~. 
E E 

CB = P -§. 


d — AC — AB + BC = w + E ); 


• • (I)- 


And 


A'C' = A'B' + BC = /(J + . 


Dividing Eq. (2) by Eq. (1): 


' = AC » 


But 


P = f e pdx = ™F e A'Cdx. 

J - e dj —e 


• • ( 2 )‘ 


If the curve DAH be assumed to be a parabola, as may be 
done without essential error, there will result: 


C A'C'dx = — ed. 
J e 3 


Hence: 


P — — we 

3 


( 3 )- 


But: 


e = V 2 Rd - d* = V 2Rd nearly. 


By inserting the value of d from Eq. (1) in the value of e , 
just determined, then placing the result in Eq. (7): 


P = 











THE RESISTANCE OF SOLID METALLIC ROLLERS. 445 

If R = R'l 


P =- R 


3 



( 5 )- 


The preceding expressions are for one unit of length. If 
the length of the roller is /, its total resistance is 


j/ 2 ^(f+g) • • • (6). 


F = PI =4 l 
3 


Or if R= R': 




In ordinary bridge practice Eq. (7) is sufficiently near for 
all cases. 

A simple expression for conical rollers may be obtained by 
using Eqs. (4) or (5). 

As shown in Fig. 2, let z be the distance, parallel to the 
axis, of any section from the apex of the cone; then consider 


!*- 

f- 




Fig. 2. 


a portion of the conical roller whose length is dz. Let R x be 
the radius of the base. The radius of the section under con¬ 
sideration will then be 














446 APPENDIX II. 

and the weight it will sustain, if R x — R ; 


dP 


l 




E + E' , 
2 w s — ^ . zaz. 


EE' 


Hence : 




E + E' 
EE' 



Eqs. (6), (7), and (8) give ultimate resistances if w is the 
ultimate intensity of resistance for the roller. 

It is to be observed that the main assumptions on which 
the investigation is based lead to an error on the sid^ 0/ 
safety. 

If for wrought iron, w — 12,000 pounds per square inch, 
and E — E' — 28,000,000 pounds, Eq. (5) gives : 


P = 


\ R \/ ? E =****' 







APPENDIX III. 


THE SCHWEDLER TRUSS. 

The general principle applied in Chapter III. to bowstring 
trusses, enables the characteristics of the Schwedler truss to 
be very simply shown. In fact, that truss is a special bow¬ 
string, having the least possible number of diagonal braces 
under the conditions assumed. 

Fig. i represents the elevation of such a truss, and the 
problem involved is the determination of such depths, near 



the ends, that one diagonal only will be needed in each 
panel; it being premised that the inclined web members or 
diagonals are to sustain tension only. 

Let W = total (upper and lower chord) panel fixed load. 

“ R r — half the fixed load (or weight) of the bridge. 
w — panel moving load. 

/ = length of span. 

d — any vertical brace or truss depth, as Cc. 
d x = vertical brace or truss depth, as Dd, adjacent to 
d and toward centre. 
p — panel length. 

a = inclination of any diagonal, as Cd } to the hori¬ 
zontal lower chord, i. e., Cdc — a. 

44 7 


u 


u 


il 


u 


u 


u 










448 


APPENDIX III. 


Let x — distance from A to the intersection of the pro¬ 
longation of any upper chord panel, in the 
left half of the truss, with the prolongation 
to the left of the lower chord. 

“ y — the normal distance from that point of inter¬ 
section to the prolongation of the diagonal 
immediately under the upper chord panel 
prolonged. 

Let the moving load pass on the bridge from A toward Z, 
and let n be the number of panel-moving loads from A. 

The reaction at A, for any position of the moving load 
will be : 


R = R l + nw • • • • (O- 

Then let the truss be imagined divided through the panel 
immediately in front of the train. 

If moments be taken about the point of intersection de¬ 
noted by x y and if T represents the tension in the diagonal 
just in front of the train*, whose lever arm is y y there will 
result : 


Ty = Rx — 



(n + i )p \ 
2 / 



Eq. (2) is so written, it is important to notice, that if the 
second member is greater than zero, or positive, T will be 
tension. Hence, if T is tension: 

Ty >_ 0 . 

But, y — (x + (n + i)p) sin a . (3). 


Also, from similar triangles ; 


d x — d 


d^ 


p ~ .r +{ n + lJp-'- X + { - n + ^ = J~ d - • ( 4 )- 


By the aid of Eq. (3) : 







THE SCHWEDLER TRUSS. 


449 


( n + i )p 

_ — R(n + i )p + n(W 4 - w) 2 R — n( W + w) 

(x + (n + i )p) sin a sin a —°* 


Hence, by the aid of Eq. (4) : 


R- 


n{ W 4- iv) 


x 4 - (» + i)p > 


R — ;/(/E + w) 


)(« + l )P 

Z---< 


pd x 


d x — d 


Using the last two members of this inequality 


i*'- 


R- 


R — n{W 4- w) 
n( W 4 - w) 




• ( 5 )- 


Or ; d > 



R — n(W 4- w) \ 
(R- *K± 2 !!iy« + I)/ • 



The second member of (6) is the least value of the depth 
d which can exist without inducing compression in the 
diagonal under consideration. This diagonal is the one 
immediately in front of the train, and the principles given in 
Chapter III. show that if this position does not induce com¬ 
pression, no other will. 

Inequality (6) shows that if R is greater than n(W 4- zv), 
or R>n(W + w), d will always be less than d v If R — 
n(W h w), then d = d x . 

Again, if R<in(W 4- w\ then will d be greater than d lf if 
tension is to be found in the diagonal. But it is not admissi¬ 
ble to make d > d x \ hence , when n becomes so great that R is 
less than n(W 4 - w), or R < n(W 4- w), T will be compression , 
and the diagonal must be counterbraced or else intcrsectmg 
diagonals must be placed in the panels , as shown in Fig. 1, near 
the middle of the truss . 

The value of n given by 


R < n(W 4- w) .(7). 

29 


1 














450 


APPENDIX III. 


will show the position of the head of the train when all 
panels between it and the centre must contain intersecting 
diagonals. All the other panels will need but one each, 
sloping upward and toward the end of the truss, as shown 

• f—> • 

in r lg. i. 

Since this method is independent of the direction of ap¬ 
proach of the train, it is only necessary to consider one-half 
of the truss. 

It is seen in (6), that d is given in terms of d lf hence the 
latter must be known in order to find d. 

The centre depth is arbitrary and may be assigned at will. 
The depths between the centre and that point indicated by 
n , in inequality (7), may also be assigned at will; consequently 
d h next to the first “ d” to be computed, will be known. The 
first “d” computed will be the “ d± for the next “ d,” etc., 
etc., to the end of the truss. 

As a margin of safety it will be well to make d a little 
greater than given by the second member of (6). 

In long spans it would be well to make the truss depth 
constant for a number of panels near the centre, perhaps, 
even between the points given by (7). This would make a con¬ 
siderable number of diagonals and panels uniform in length, 
which would otherwise lack uniformity. Thus the construc¬ 
tion would be simplified and cheapened. 

The loads have been taken uniformly, but precisely the 
same methods would hold if they were not uniform. 


Example. 

Let the following example (the truss shown in Fig. 1) be 
taken : 

Span = l = gp = 108 feet; .\ p — 12 feet. 

Centre depth = 16 feet. 

W — 8.00 tons. w= 18.00 tons. 

(W + zu) = 26.00 tons. 

R 1 = 4 W = 32.00 “ 


THE SCHWEDLER TRUSS . ^ 

From Eq. (i): 

R = 32 + i%(n - - >r J 8 n ) . . . (8). 

If » = 3, by (8) and (7): 

R = 32 + 18 (3 - |) = 74 < n(W + w) =; 78. 

Hence the diagonals De and AV/, in the panel in front of 
the head of the train, at d , must both be introduced. 

If n = 2, by (8) and (7): 

R = 62 > n(W + w) — 52. 

Hence Cd is the only diagonal needed in the panel CDdc , 
and d — Cc is to be computed from Eq. (6). 

The centre depth — Ee — Ff was taken at 16 feet; let 
Dd be taken at 15.5 feet. 

Since n — 2; R — n (W + w) = 10, and — ; / (W + w ) _ ^ 

2 

Substituting these values, and = 15.5, in Eq. (6) : 
^=0.91 d t = 14.11 feet. Hence let 
= 14.5 feet = dV (Fig. 1). 

Next, let the head of the train be at b, i. e., let n = 1, 
Then by Eq. (8): 


R — 32 + 16 — 48. 

Also ; R — n( W + w) = 22, and R — ^ ^^ = 35. 
For this position of load, d t = 14.5 feet. Hence, by Eq. (6): 
d = 14.5 ^1 — = 9.94 feet. Hence let 


d = 10.00 feet = Bb (Fig. 1). 





452 


APPENDIX III. 


Fig. i represents the truss, drawn to scale, with the various 
depths given or computed as above, i. e ., 

Ec — Ef = 16.0 feet. 

Dd— Gg= 15.5 “ 

Cc = Hh — 14.5 “ 

Bb = Kk = 10.0 u 

In the three panels adjacent to each end of the truss only 
two main diagonals are thus seen to be necessary, and in 
those no compression will ever exist. In each of the three 
middle panels, however, two intersecting diagonals will be 
necessary, since no diagonal must sustain compression. 

As is evident, the expression R — n( W + w) is the vertical 
shear at the head of the train. Hence the limiting case of 
the inequality (7): 


R = n( W +w), 

gives the two points, in the two halves of the truss, at which 
the vertical shear at the head of the train is zero. Between 
these points intersecting diagonals or counterbraces are 
needed, and only between therm 

After the truss depths are fixed by the preceding method, 
the stresses in the individual members are to be found in the 
usual manner—as in any other bowstring truss—as shown in 
Chapter III. 


APPENDIX IV. 


REACTIONS AND MOMENTS FOR CONTINUOUS BEAMS. 


Case I.—Three Spans with Two Intermediate Points of Support and 

Two End Supports. 

The notation of Art. 35 will be used and reference will be 
made to Fig. 1 of that Art. 

M, - [/,./>(, - - *■« + /,) ip{, - i )|] + 

f4 (4 + 4) (4 + 4) ~ .(0- 

M 2 =- { M S L + lfsp{i - ~)j }+2(A + 4 )- • • (2). 


+ f.<-"• 

R, = 2Pj Ml .(4). 

r, M 2 -M z M z 

K 3 = - 7 - + 2 :Fj - J- .( 5 ). 

^2 *3 ‘3 

«=M'“D + f. w 


In ordinary swing-bridges where 4 = 4 = l and a single 


weight P rests on 4: 



^ \ % 2/ (/ 4 - 4 ) ) / x 

/*// * 4 (/ + 4) 8 -4 a f * * (7> 



z Ik 

7 * 4 (/ 4 - 4) 2 - 4 s 















454 


APPENDIX IV. 


These formulae are in no wise changed for a single weight 
P resting on 4 except that R x and R± are interchanged. Also : 


r> (^l — l — P (l ~ Z) r> 

-^3 — ~l 



.*. R 2 = P — R\ — ^41 ^3 



Case II.—Two Spans with One Intermediate Support. 

Reference will be made to the notation and Fig. 2 of Art. 35. 

m .=--99 

. <■*>• 

R, = 2Pj - ^ + 2P*- - ^ .(13). 

4 4 4 4 

-£> + f. (,4> ' 

If 4 = 4== l: 

-?)/-!■ • - <,s > 


These formulae are based on the supposition that there 
may be negative reactions at A and C of Fig. 2, Art. 35. If 
no negative reactions are possible, and if the load is on one 
arm only, that arm will be a non-continuous beam for such 
load, and the reactions will be found by the simple principle 
of the lever. 










APPENDIX V. 


CANTILEVERS. 

Art. I. Cantilever Structures.—Positions of Loading for Greatest 

Stresses in the Cantilever Arm. 

Cantilever structures are formed of continuous or semi- 
continuous trusses in which the reactions and stresses are 
equally determinate with those in non-continuous trusses. 
Fig. I, PI. XIII., typifies a cantilever structure with an 
anchor arm at each end. The anchorage at A holds the 
anchor arm n in position under all conditions of loading, thus 
enabling the cantilever arm m to sustain the suspended span 
/. The latter is a simple, non-continuous truss with inclined 
end-posts, suspended at the lower extremities of the vertical 
tension members at C and D } thus readily allowing expansion 
and contraction to take place both in the suspended span 
itself and in the cantilever arms adjacent to it. As cam 
tilevers are erected without lower false-works between the 
piers B and E, each cantilever arm, with the adjacent half of 
the suspended span, is built out from each of those main 
piers ; hence the members FG sustain compression during 
erection, while HD take tension. These erection stresses 
are frequently very heavy, and the members themselves must 
be designed to take them.* At the same time, proper details 
must be arranged so that after the structure is complete the 
requisite expansion and contraction can take place at each 
end of the suspended span. These results are usually accom¬ 
plished by folding wedges, or wedges and rollers, removed 
after erection, in connection with oblong pin holes. 

Similarly, nearly the whole of the upper chord of the sus¬ 
pended span must be designed to take the tensile erection 
stresses, and nearly the whole of the lower chord the corre¬ 
sponding compression erection stresses. The erection stresses 
in the cantilever arms are, of course, the same in kind as 


45 6 


APPENDIX V. 




those caused by the moving and fixed loads, and no reversion 
takes place. 

The conditions of loading for the greatest stresses in the 
suspended span, in addition to the erection stresses, are pre¬ 
cisely the same as those for the ordinary non-continuous 
truss given in detail in Art. 7, and they will receive no fur¬ 
ther attention here. 


Greatest Web Stresses in Cantilever Arm. 


In seeking the greatest web stresses in the cantilever arm, 
reference will be made to Fig. 1, PI. XIII., and a perfectly 
general system of loading will be assumed. The chords 
of the cantilever arm will also be assumed to be not parallel. 
Let the greatest stress X, in MN, be desired, and let the 
system of weights or concentrated loads W ly W 2 , etc., extend 
over the entire suspended span /, and over the cantilever arm 
m, from C to some point to the left of M. Then let i rep¬ 
resent the distance from C to the point of intersection of the 
chord sections in the same panel with MN, while h is the 
length of the normal dropped from the intersection point to 
MN prolonged. Also let g x represent the distance from 
D to the centre of gravity of the load on /; g 2 the distance 
from the same point to the centre of gravity of the load to 
the left of C; g s the similar distance from M for the load 
to the left of M\ and g 4 the similar distance from M of the 
loads W 4 , W 5 , etc., in the panel in question. The loads on 
the left of M are W x , W 2 , etc. . The distance from C to M 
is represented by k. R is the downward reaction or upward 
pull on the anchorage at the extremity of n ; while R x is the 
upward reaction at the main pier. The weights over the 
various portions of the structure will be represented as 
follows: 

1^ 

^ W for total weights in /. 


2 W 

k 

2W 

2 W 


it tt 


It it 


it 


ft 

tt 

m. 

it 

it 

k. 

tt 

tt 

1 / 


“ “ m + l. 


it 


CAN TILE VERS. 


457 


The centre of gravity distances, g u g 2} and g 3 , can be read- 
ily found by aid of tabulations like those given in Arts. 9 and 
11, whiles will be found by taking moments of W 4 , W 5 , etc., 
about M. 

Then, by moments in the ordinary manner: 

R .(,). 


R 2 = reaction at C from load on suspended span = ~ 2 W (2). 

4 


R, = R + R,+ 21V = W+ - + i) 2 W (3). 

If the length of the panel in question is/, then that portion 
of the loads W 4 , W 5 , etc., resting in it and carried to M, is: 

(W t + W, + etc.) (1 -fj .(4). 

If the cantilever arm be now supposed divided through 
the panel MN, and if the moments about intersection of 
chords be taken of all the forces external to that portion of 
the structure to the left of this line of division, including 
the reactions R and — R lf there will result: 


S = lA?f 2 lV+(&-l+i) 2 lV 


— (W x + W 2 + etc. + W 4 + W 5 + etc.) (k + i) 

— (Wt + W 2 + etc.) g 3 + (IV 4 + W 5 + etc.)(£ 4 - t)^ |. . (5). 

In order that 5 may be a maximum or a minimum, dS 
must be zero when g u g 2 , g 3 , and g 4 each vary by the same 
small amount, as when the loads are all advanced slightly to 
the left. It is to be remembered, however, that the variation 
of g x will be opposite in sign to that of the others. By thus 






458 


APPENDIX V. 


making AS equal to zero, there will result as the desired con¬ 
dition : 

w i+ W 2 + etc. + ( W .i + W, + etc.) = *-SW + 2 IV.- 

.-. ( W t + W s + etc.) A±A = * SW+ 2 W. . (6). 

In case there may be a number of maxima indicated by 
Eq. (6), the greatest must be sought by trial. Eq. (6) shows, 
as might have been anticipated, that the loads W 1 + W 2 + e tc., 
between the panel in question and the main pier, do not 
, directly affect the conditions for greatest stress. 

If the load is of the uniform intensity w, and x is the por¬ 
tion of the panel p covered by it, Eq. (6) is only satisfied by 
making x — p. This result might also have easily been 
anticipated, for it is clear-that if the load is uniform it must 
at least reach to M, for the panel MN. The amount by 
which it may extend to the left of M is a matter of indiffer¬ 
ence. 

If the chords are parallel , i is infinitely great and Eq. (6) 
becomes : 

W 4 + W 5 + etc. . . . (7). 

In the general case, when Eq. (6) has fixed the proper 
position of loading, the corresponding values of g\, g 2 , g 3 , and 

can be at once obtained by the methods already indicated, 
and Eq. (5) will then give the desired greatest value of 5 . In 
the case of a uniform load, w, that equation takes a much 
simplified form, since: 

W x + W 2 + etc. = o ; W x + W 5 + etc. = wp ; 

m l 

2 W=wk; ^ W= wl ; 

l r , k p 

gi = ~ > Si = 1 t- : £3 = 0, and^j = A 





CANTILEVERS, 


459 


Substituting these values in Eq. (5) : 

S = Th {*('+*) + (*-/)(* +*’)} .(8)- 

If the chords are parallel and a is the angle between £ and 
a vertical line,// = (00 = /) cos a ; hence: 

5 = w + k — ^ sec a .(9). 

With the same uniform load, w , per lineal foot, and with 
kiv on the cantilever arm m, Eqs. (1), (2), and (3) become: 

R — w — ( ~ + k — •— ).(10). 

n \2 2mJ 

„ Wl r v 

^'2=-.(II)- 

R l = ^(™+ l) + W k(™-^ + l) .(12). 

1 2 \n J \n 2n J 

If the load, w, covers the entire arm m, k = ///, and : 

R — w ~ {l in) .. ' . > c ...(13). 

2;/ 

^ = y, and R t = y (* + i) + (£ + 1 )• <H). 

If a single weight, W, rests on the suspended span, /, at the 
distance g x from D : 

= ■ OS)- 

If a single weight, W } rests on the arm m at the distance k 
from C: 

r = W(m - k) ^ _ Q. and7?i = w + i\ (16). 

n \ n J 











460 


APPENDIX V. 


Greatest Chord Stresses in the Cantilever Arm. 

Reference will again be made to Fig. 1 of PI. XIII., in 
which the compression web members are vertical; the nota¬ 
tion will remain as before. The greatest chord stress, S c , in 
any panel, as MN, will be sought. Let h l represent the 
normal dropped from M on the chord panel whose greatest 
stress is S c ; then, if moments be taken about M, h y will be 
the lever arm of S c , and there will result: 

S c = Jj | R x {m — k) — R (m + n — k) 

- (W x + W 2 + etc.)£ 8 1 . . . . (17). 

/. S 0 = i W-^p2lV-(/+ k)2W 

— (W t 4- W 2 4- etc.) ^3, | • • • • (ic). 

For a maximum or minimum AS C — o when g 2 , and are 
the only variables in the second member of the equation ; 
hence: 


- W= 2 W-(IV, + W 2 + etc.) = 2 W. . ' . (19). 

This condition for a maximum is seen to be independent of 
/; hence it is precisely the same whether the chords are 
parallel or inclined. After the values of g u g 2 , and g 3y corre¬ 
sponding to the position of loading fixed by Eq. (19), are in¬ 
serted in Eq. (18), the desired chord stress will at once result. 
It is to be borne in mind, however, that there may be several 
maxima fixed by Eq. (19), and that the greatest of these, to 
be found by trial, is the “greatest stress” desired. 

If the load is of the uniform amount, w, per lineal unit of 

k 

structure, Eq. (19) is satisfied only by making 2 W — kzv, 
hence, the movhig load must cover the suspended span, and the 
cantilever arm from its end to the vertical line through the cen - 


CAN TILE VERS. Af 1 

tre of moments — i.e ., the distance k. Eq. (18) then takes the 
value, for uniform loading : 


•Sc = - ^ (k + l) .(20). 

For a panel in the horizontal chord, h x is simply the depth 
of truss at the origin of moments. 

In case all zveb members are inclined , as indicated in Fig. i, 
so that any lower chord panel point M x is at the horizontal 
distance q from that in the upper chord, Eq. (18) takes the 
form : 

= Ji | g&W - + f^ 1 ZlV-(l+ k + q)2W 

- (IV, + IV 2 + etc.) (/ - q) g A . . . . (21). 



Again making 4 S C = o, with g u g 2f and g s the only vari¬ 
ables, there results for a maximum or minimum : 

2 W = ^±l 2 W + {IV, + W 2 + etcg-gl. . . (22). 

/ p 

Or, 

k ±A 2 W = 2 W-( IV, + W 2 + etc.)I— 2 ' . . (23). 
I P 


If q — o, Eq. (19) at once results. The observations fol¬ 
lowing Eq. (19), regarding i, g x ,g 2 , and g 3 , obviously hold true 
for this case also. If the moving load is of the uniform in* 












462 


APPENDIX V. 


tensity w, Eq. (23) is satisfied only by making it covert + p\ 
hence that equation takes the form : 


k + q = k + 


P 


. . (24)* 


which shows that the uniform moving load must cover the sus¬ 
pended span , and the cantilever arm from its end to the further 
extremity of the panel in which the chord stress is sought. 

For this uniform moving load Eq. (21) takes the form : 

s ‘ = ji | ^ * — ~{k + q) (k + p + 0 - O - q) p ~ 2 1 • (25)- 

If q — o in this equation, Eq. (20) immediately follows. 

If the panel in the lower chord is under consideration, M, 
Fig. 1, becomes the origin of moment, and as the moving 
load is supposed to traverse the upper chord, the conditions 
for a maximum and the corresponding formulae are precisely 
the same as for the case with vertical compression web mem¬ 
bers. 

If, with all inclined web members, the moving load tra¬ 
verses the lower chord, the preceding conditions and formulae 
apply precisely as they stand. It is only to be borne in 
mind that the formulae involving q are to be used for the 
chord carrying the moving load. 

If the tension members are vertical , and the compression 
members inclined , as in the Howe truss, or as exemplified by 
the inclined post, P, Fig. 1, of PL XIII., the positions of 
moving load for greatest chord stresses will be at once found 
by making q — p in Eq. (23), which gives: 

k -\~ p l m 

—J- 2 W= ZW .(26). 

for the general load, and the same condition for a uniform 
load as that described in italics immediately after Eq. (24). 
By making q — p, in Eq. (25), the expression for the greatest 
chord stress under a uniform load, w, becomes— 




CANTILEVERS. 


463 



w \ (k + py (k + p)l 
h' | 5 + ^ 


(k 4 - p)w 
2/1 1 


(k+p + l) (27). 


If the truss is of uniform depth, Ji 1 is that depth. 

The preceding investigations hold true for any system of 
triangulation or for any system of loading. They do not 
apply to multiple systems of bracing, as such systems do not 
admit of exact calculation of stresses. When they are used, 
each system must be assumed to act independently of all the 
others (although the assumption cannot be definitely estab¬ 
lished) and to carry the system of maxima concentrations 
permitted by the moving load used, or else such a uniform 
load as will be practically equivalent to the real load. 

Although there is not at the present time (1890) any con¬ 
structive or other proper reason for the use of multiple sys¬ 
tems of bracing in any pin structure of proportions hitherto 
employed, the above assumptions are within the limits of 
safety, and may be used where necessary and unavoidable. 

Another constructive device is that -shown in Fig. 2. It 
may be used advantageously in cases where the cantilevers 
are supported on high iron or steel piers or towers, when the 
latter would become too expensive if constructed of masonry 
of corresponding dimensions. It leads to appreciable econ¬ 
omy by the shortening of the clear cantilever opening, as 
well as the anchor arm. It consists in separating the feet of 
the two inclined posts, P and IB', of Fig. 1, PL XIII., by some 
convenient distance, 0, and omitting all bracing in the rect¬ 
angle, BB'B^B^ Fig. 2, so that no shear can be transferred 
past B' to the left, or past B^ to the right. Since this last 
condition exists, the reaction, R , and the positions of loading 
for all the maxima web and chord stresses in the cantilever 
arm, as given by the preceding formulae, will hold for this 
case without any change whatever. It is only to be carefully 
observed that m and n are to be measured from the extremi¬ 
ties of the cantilever and anchor arms, respectively, to the 
extremities of the open panel, 0, as shown in Fig. 2. The 
reaction, R / at B\ will equal the sum of the vertical compo¬ 
nents of all the inclined stresses in the first panel of m — i.e 





464 


APPENDIX V. 


the total shear in that panel, added to the panel loads acting 
at B and B'. Similarly, the reaction R± at B z ' will equal the 
sum of the vertical components of all the inclined stresses 
in the first panel of n — i.e ., the total shear in that panel, 
added to the panel loads acting at B l and BJ. Finally, the 
reactions, R z ' and R", of Fig. 2, added together, will equal 




the reaction, R„ of Fig. 1, PL XIII. Thus it is seen that 
the formulae already established for the latter case will meet 
the former without any change whatever. 

The case of the subdivided panel shown in Fig. 3 is also 
covered by the preceding formulae. The general and detailed 
considerations governing their applications are precisely the 
same as those given at length in Art. 7 ; it is not necessary, 
therefore, to give them further attention here, except to ob¬ 
serve that for the greatest stress in be , the panel length to be 
taken is ad , while ae is the panel for the greatest stress in ab. 

Art. 2.—Positions of Moving Load for Greatest Stresses in the Anchor 
Arm, and the Greatest Stresses Themselves. 

No load placed on the anchor arm of a cantilever structure 
will affect in any way the reaction or any shear in the canti¬ 
lever arm, except such as may result from its slight deflectional 
movement, being too small to be expressed in appreciable 
amounts. It at once follows from this fact that all stresses 
in the anchor arm, due to loads on that arm, may be compu¬ 
ted precisely as if it ( AIB'L of Fig. 1, PI. XIII.) were a simple, 
non-continuous truss supported at each end (i.e., A and B l 






















CAN TILE VERS. 


465 


of Fig. 1, PI. XIII.). The greatest of these stresses so found 
are then to be combined with the greatest stresses of the 
same kind due to the load on the cantilever arm and result¬ 
ing from the reaction, R, at the anchorage ; this combination 
will give the resultant greatest stresses desired. 

Although the preceding general observations are compre¬ 
hensive and sufficient for the determination of all the greatest 
stresses desired, it will be well to extend them in some detail, 
without, however, establishing any formulae. 

Inasmuch as all upper chord stresses in the anchor arm, due 
to load on that arm only, will be compressive, it is clear that 
the greatest tension in the upper chord, so far as it may 
exist, will be found with no moving load on the anchor arm, 
and with the moving load so placed on the cantilever arm as 
to give the maximum bending moment (and, hence, chord 
stresses) over the main pier between the anchor and cantilever 
arms. This position of the moving load on the cantilever 
arm can be found for the various cases from Eqs. (18), (20), 
(21), and (25) of the preceding Art. The resulting moment 
over the main pier, divided by n, will give the reaction, R, with 
which the greatest tension throughout the upper chord and 
the greatest compression throughout the lower chord are to 
be found— i.e., so far as this tension or compression exists. 

If the anchor arm is very long in comparison with the can¬ 
tilever arm, the fixed load compression in the upper chord 
and tension in the lower may be so great that there will be 
but little upper chord tension and lower chord compression, 
even for the conditions producing maxima. 

Inasmuch as all loads on the suspended span and cantilever 
arms produce tension in the upper chords of the anchor arms 
and compression in the lower, the positions of moving load 
for the greatest compressions in the upper chord of either 
anchor arm, or greatest tensions in the lower, are precisely 
the same as if it (the anchor arm) were a simple, non-continu- 
ous truss, the cantilever arms and suspended span, at the 
same time, carrying no moving load. The greatest compres¬ 
sive stresses in the upper chord and tensile stresses in the 
lower chord, thus found in the anchor arm, as for a simple, 


1 


466 


APPENDIX V. 


non-continuous span, are to be combined with the upper 
chord tension and lower chord compression produced by the 
fixed load of the cantilever arms and suspended span, whence 
will result the maxima chord stresses desired. 

The greatest shears in the anchor arm will evidently be 
affected by the magnitude of the reaction R, Fig. 1, PI. XIII., 
at its extremity. If a downward shear, producing tension in 
members sloping similarly to OQ, or compression in IB', is 
called a main shear, then any condition of loading on the 
structure which will increase the downward reaction, R, will 
increase the main shears and, hence, the stresses in the main 
web members, as AL, LV } UT, etc., and IB' will be called. 
But the maximum value of the downward moving load reac¬ 
tion or pull at the extremity of the anchor arm will be 
found with no moving load on the latter, and for the maxi¬ 
mum value of the bending moment over the main pier B', as 
given by Eqs. (18), (20), (21), and (25), of the preceding Art., 
as already stated on page 465. Finally, that part of the 
maximum main shear in any panel of the anchor arm due to 
the moving load on that portion of the structure alone, is 
found precisely as if it were a simple, non-continuous truss 
with the given system of loading, advancing from the extrem¬ 
ity of the anchor arm to the main pier. The positions of 
moving load, therefore, for the main shears or main web 
stresses are, first, so much on the cantilever arms and sus¬ 
pended span as will produce the maximum downward pull at 
the extremity of the anchor arm, and, second, with this con¬ 
stant condition in the main span of the structure, an advanc¬ 
ing moving load from the extremity of the anchor arm, 
treated as a simple, non-continuous span, to the main pier. 
The maximum moving-load shears or stresses thus found, com¬ 
bined with the fixed-load shears or stresses in the anchor arm, 
will give the resultant greatest web member stresses desired. 

It may sometimes happen that the anchor arm will be so 
long that the greatest downward pull at its extremity due to 
the moving load will be less than an existing upward reac¬ 
tion due to the fixed load. In such a case no anchorage, of 
course, will be required, as the reaction at the extremity of 


CAN TILE VERS . 


467 


the anchor arm will always be upward. The preceding con¬ 
ditions for greatest main web stresses, however, hold in all 
cases without exception. 

The greatest counter shears and, hence, stresses in counter 
web members such as QV, or j VL if under compression, or 
A.L, in tension, Fig. 1, PI. XIII., will, for the same general rea¬ 
sons just stated in connection with the greatest main shears, be 
found under the moving load advancing from the main pier 
B to the extremity of the anchor arm, under the supposition 
that the latter is a simple, non-continuous truss, with no mov¬ 
ing load whatever on the cantilever arms or suspended span. 

If a portion of the web members are vertical posts, as 
shown in Fig. 1 of the plate, so that the kind of stress in 
them is the same whether acting as counters or main web 
members, those nearest the extremity of the anchor arm will 
probably take their greatest stresses as counters, but those 
more remote are doubtful and must be computed both as 
main and counter members in order to find the greatest 
values. Those near the main pier will take their greatest 
stresses as main web members. 

Art. 3.—Positions of Moving Load for Greatest Stresses in Anchor 
Spans and the Greatest Stresses Themselves. 

The anchor span is shown as the span n in F'ig. 2, of PI. 
XIII., on the left of which there is supposed to be a cantilever 
arm similar to m on the right. The anchor span is always of 
such length and weight that the reactions, R and R u are invari¬ 
ably upward. Hence the anchor span is identical in character 
with the anchor arm, whose length and weight are so great 
that the reaction at its extremity over the anchorage is 
always upward. Hence the positions of moving load for the 
greatest stresses are precisely the same as those established 
for the anchor arm in the last Art. 

The greatest moving-load stresses in the span n, Fig. 2, PI. 
XIII., are to be determined precisely as if the trusses were of 
a simple, non-continuous character ; and the fixed-load stresses 
are to be found in the same trusses under exactly the same 
conditions. The greatest difference of moments at the two 


468 


APPENDIX V. 


sections, BR X and AR, is then to be found and divided by the 
span length n, thus giving a reaction which is to be treated pre¬ 
cisely as the downward reaction at the extremity of the anchor 
arm in the preceding article. This downward reaction will pro¬ 
duce stresses in the chord and web members of the anchor 
span, which are next to be determined in the usual manner for 
a load hanging at the extremity of the span n, with the other 
extremity fixedly held. If both cantilever arms flanking the 
span n are of the same length m , and of the same weight 
similarly disposed, the fixed-load moments at the sections 
AR and BR X will always be equal (but not otherwise), and 
the difference in moments mentioned above will be entirely 
due to the moving load on one of the adjoining cantilever 
spans. Any difference of moments at those two sections 
resulting from dissimilarity or inequality of fixed loads must 
be added to or subtracted from the unbalanced moving-load 
moment just described. Finally, the fixed-load balanced mo¬ 
ment at either of the sections AR or BR X is then to be deter¬ 
mined and divided by the depth of the truss in order to find 
the uniform tension throughout the length of the upper 
chord, and the uniform compression throughout the length 
of the lower, if the truss has a uniform depth. The stresses 
resulting from these four sources—viz., the moving load and 
the fixed load on the span n\ the unbalanced moment over 
the piers due to possibly both moving and fixed loads, and 
the balanced moment due to fixed load only—are to be so 
combined as to produce maxima in each and all the truss 
members. 

The greatest web stresses will require the greatest unbal¬ 
anced moving-load moment alternately at each end of the 
anchor span, with the usual progressive movement of moving 
load for ordinary non-continuous trusses; but the greatest 
upper chord compression and lower chord tension will be 
found with no moving load on the adjacent cantilever arms 
or suspended spans. 

The greatest upper chord tension and lower chord com¬ 
pression will be found by combining the fixed-load stresses, 
found as indicated above, with the moving-load stresses 


CAN TILE VERS. 


469 


induced by the greatest possible moving-load moments 
simultaneously at each end of the anchor span, the latter, at 
the same time, being entirely free from moving load. 

If the anchor span is not of uniform depth, the balanced 
fixed-load moments at its extremities will induce stresses 
in the web members, which are to be determined in addition 
to those in the chords for combination with the other 
stresses described above in the search for the maxima. 

Art. 4.—Wind Stresses. 

The stresses due to the action of the wind on the various 
portions of a cantilever structure play a very important part 
in its design. The conditions of wind loading are much more 
varied than in ordinary non-continuous spans, and the result¬ 
ing stresses must be computed with great care and thorough¬ 
ness. The pressures against the different members of the 
structure constitute a fixed load, and all resulting stresses are 
to be found by precisely the same general methods as given 
in the preceding articles for the vertical fixed loads— i.et, own 
weights. 

Fig. 5 of PI. XIII. shows the horizontal system of lateral 
bracing between the horizontal upper chords, ABODE of 
Fig. 1. If the wind blows on the entire structure in a given 
direction, as shown by the arrow in Fig. 5, there will in gen¬ 
eral be a horizontal reaction induced at A, in direction and 

1 

amount depending on the relative proportions of anchor and 
cantilever arms and suspended truss. This reaction must be 
provided for, at A , by a proper connection between the end 
of the anchor arm and masonry at that point, so designed 
that the requisite longitudinal expansion and contraction 
may at the same time take place. The amount of this reac¬ 
tion is to be determined precisely as the fixed-weight reac¬ 
tion, R, at the same point, in Fig. 1. This reaction, R , will 
be composed of two parts, one of which is due to the wind 
pressures along the upper chord, ABCD, Fig. 1, and the other 
to those along the lower chord, ALB'NG; and each is to be 
found as indicated above. By means of these horizontal 
reactions at A , and the panel wind pressures at the upper 


4/0 


APPENDIX V. 


and lower chord points, all the fixed-load wind stresses in 
the upper and lower lateral systems, both of which are pro¬ 
jected in Fig. 5, may be at once completely established. 

The wind pressures against the moving train form a mov¬ 
ing load in those chords carrying the train (the upper chords 
in Fig. 1, PI. XIII., and the intermediate and lower chords in 
Fig. 3). The conditions under which these moving wind 
loads produce their horizontal reactions at A, and their great¬ 
est stresses in the web and chord members of the lateral sys¬ 
tem, are precisely the same as those described in detail for 
the vertical or train moving loads in the preceding Arts. 
The cantilever conditions, so to speak, including span lengths 
for the former loads, are identical with those for the latter. 
Hence the positions for the wind maxima must be deter¬ 
mined by exactly the same general methods as are used for 
the vertical loads. The computations for the moving wind 
stresses are simplified by the fact that that load is of uni¬ 
form intensity. 

The combination of the fixed and moving wind load 
stresses according to the usual methods will give the desired 
resultants in all the members. 

In the case of the anchor span of Fig. 2, PI. XIII., the same 
general observations as those already made are to be applied. 
The fixed wind loads on the span are to be treated precisely 
like the fixed vertical loads, and the resulting wind stresses 
in the upper and lower lateral systems are to be computed 
accordingly. Again, the moving wind loads accompany the 
moving vertical or train loads, and all the conditions deter¬ 
mining the greatest stresses for the latter determine those 
for the former also. The same set of greatest moving wind 
stresses must therefore be found as for the moving vertical 
loads, and followed by their combination with the fixed wind 
stresses, in order to reach the resultant wind stresses desired. 

Inasmuch as it is known that the highest wind pressures 
cover comparatively small areas, it will usually be necessary 
to supplement the preceding wind computations, which are 
based on the assumption that the wind presses equally over 
the entire structure, by others resulting from the application 


CANTILE VERS. 


471 


of the highest pressure to some particular portion, as the 
clear cantilever span or the anchor span, or, possibly, the 
anchor arm, and provide for the resulting stresses. 

The extent to which these supplementary and special 
computations are to be made will depend upon the judgment 
of the engineer acting on the circumstances of each case. 

The transverse bracing in the vertical and inclined planes 
of the various pairs of web members is to be designed under 
the same assumptions of wind transference, and in accord¬ 
ance with the same principles used for the same general 
purposes in ordinary non-continuous spans. The wind loads 
in the upper chords of the suspended trusses of the cantilever 
span should always be carried down to the lower chords of 
the cantilever arms at their extremities, and alone those 
chords to the piers, in order that the overturning wind effect 
on the cantilever and anchor arms may be reduced to a mini¬ 
mum. In the same manner, the wind loads in the upper 
chords of the cantilever and anchor arms, which are resisted 
by the main piers, should be carried down to them by the 
transverse bracing between the inclined posts acting as por¬ 
tals at those places. In high cantilever trusses the overturn¬ 
ing wind effect is frequently a very serious matter, and should, 
indeed, be carefully computed in all cases, and its results com¬ 
bined with the fixed load and wind stresses already deter¬ 
mined. It will throw a considerable portion of the fixed load 
in the windward truss into the leeward. It will also consid¬ 
erably increase the downward pull on the windward side of the 
anchorage, and relieve that at the leeward side by the same 
amount. These changes in the downward pull of one side of 
the anchorage can be found by dividing the wind moment 
about the top of the pier between the anchor and cantilever 
arms by the transverse width between the trusses where they 
are attached to the anchorage. The same effects occur, of 
course, similarly in the anchor span. These overturning 
effects, resulting from the pressure of the wind against the 
structure, affect the fixed-load stresses only. It may also be 
necessary to consider the truss stresses arising from the over¬ 
turning effect of the wind on the train. This, however, will 


472 


APPENDIX V. 


depend upon the special circumstances affecting any particu¬ 
lar case. 

Art. 5.—Economic Lengths of Spans and Arms. 

The problem of the relative lengths of suspended spans, 
cantilever arms, anchor arms, and anchor spans for the most 
economical amounts of material in them is not capable of 
exact mathematical treatment, but may be very simply solved 
in a manner sufficiently accurate for all practical purposes. 
In order to do this it is only necessary to observe that the 
weight per lineal foot of any single span of bridge structure, 
omitting the floor, is very closely found by multiplying the 
span length by a numerical factor, determined by actual com¬ 
putation of weights of similar spans. In the case of a canti¬ 
lever arm, the length of that arm would be used instead of 
span length. In the analysis which follows, there will be used : 

a — cantilever arm factor ; b = suspended span factor ; 

c — anchor “ “ ; d — anchor “ “ 

L — length of cantilever span = l + 2 m of Fig. i, PI. XIII. 

/ — “ suspended span. 

Economic Length of Suspended Spent and Cantilever Arm . 

The total weight, W, of the suspended span and cantilever 
arm trusses will be : 

\(L- If + bl 1 - W. 

Therefore, by differentiation : 

a(L — l) - 2bl — o. :.l— — L. . . . (i). 

a + 2 b x } 

l, 

Hence, cantilever arm = m —- L .(2^ 

a + 2b v J 

If there be taken loosely, a = 20 and b = 7, / will be nearly 
0.6 L and m nearly 0.2 L. It has been found, however, that 
/ may be advantageously taken about 0.5 L to 0.55 Z, and tn 
about 0.25 L. 





CAN TILE VERS. 


473 


Economic Length of Anchor Arm. 

There is to be found in this case the entire weight of the 
trusses for two anchor arms and the cantilever opening, or 
the weight of / + 2n + 2m — S of Fig. i, PI. XIII. By using 
the proportions established in the preceding case, there will 
result : 

Anchor arm length == n. 

Suspended span length = — a — J (S — 2n) = l. 

Cl -j - 2u 


Cantilever arm length =- 

& a + 2b 


(S — 2n) = m. 


Hence the entire weight desired will be : 

ab 


2cn 2 + - ■ (S - 2n) 2 = W. 
a + 2b J 


• • 


( 3 )- 


By differentiation: 
ab 


cn — 


a —j— 2 b 


(.S — 2n) = o. n — 


ab 


c (a + 2b) + 2 ab 


.\l = 


ac 


c(a + 2b) + 2 ab 


S ; and, m 


be 


c (a + 2b) + 2ab 


S. ( 4 ). 


5 . ( 5 ). 


If, as before, there be taken loosely: 

a — 20; b — y ; and c — io, 

there will result: 

n — 0.226 S ; m = o. 113 S, and / = 0.32 S'. 


It has been found in the best practice that the anchor arm 
should be 1.67 to 2 times the length of the cantilever arm, 
with the suspended span about twice the length of the latter, 
or a very little more. 


Economic Length of the Anchor Span. 

The economic length of the anchor span will be found by 
considering the total weight of the trusses for an anchor span 









474 


APPENDIX V. 


and an adjoining cantilever span, or the weight of n + 2 m + l 
in Fig. 2, PI. XIII. 

The proportions established in the preceding cases give : 
Anchor span length = n. 

Cantilever arm “ = —-— 7 (S — n). 

a + 2 b 


Suspended span 


a 

a + 2b 


(S - 7 l). 


Hence the total weight desired is: 


dir + 


ab 

a + 2 b 


(5 - nj = 


W. 



Differentiation then gives: 


n 


d + 


ab 


ab 


a + 2 bj a A- 2b 


S. 


n — 


ab 


d (a -f 2 b) + ab 


S. ( 7 ). 


Hence: 


/ 


ad 


d(a + 2 b) + ab 


S\ and m = 


bd 


d (a + 2 b) + ab 


S. . (8). 


If the same rough factors as before be taken, viz., a — 20; 
b — 7, and d — 12, there will result: 


n =0.26 5 ; / = 0.4385; and m = 0.153 S. 

These results, however, make n too small for the ordinary 
requirements of navigation where such exist. The strict 
economic requirements in such cases are, therefore, neglected, 
and n taken from 0.3 to 0.4 5 . These results show the 
importance, however, of making the anchor span, compara¬ 
tively speaking, very short. 


Art. 6.—Wind Pressure. 

The great importance of wind stresses in cantilever struct¬ 
ures makes it of much interest to observe that the latest 











CANTILEVERS. 


475 

i 

investigations indicate a somewhat lower maximum on large 
areas for the usual highest winds than has hitherto been 
contemplated. The highest pressure actually observed and 
measured at a very exposed point on the site of the Forth 
bridge, over a period of six years, on a surface of 300 (15 ft. 
by 20 ft.) square feet, was 27 pounds per square foot, 
although 41 pounds per square foot was at the same time 
observed on a small area of 1.5 square feet immediately 
adjacent to the large one. A pressure of 50 pounds per 
square foot, therefore, over the entire surfaces exposed in a 
500 feet cantilever span is probably as rare at any given 
location as a cylone at the same place. 

Again, some later investigations of a most valuable char¬ 
acter, by Mr. O. T. Crosby, and given in his paper, “An 
Experimental Study of Atmospheric Resistance," read before 
the West Point Branch of the United States Military Service 
Institution, in 1890, appear to quite invalidate the formula, 
V 2 

P— -- in which P is the pressure in pounds per square 

100 

foot, and V the wind velocity in miles per hour, given in 
Art. 80, page 370. Mr. Crosby’s experiments with a surface 
one square foot in area show that with velocities varying 
from 30 to 130 miles per hour, very closely : 



This is such a radical departure from the hitherto accepted 
values for P that further tests are desirable, although it is 
difficult to detect any ground of error in Mr. Crosby’s 
results. 




i 





t 




/ 




Pi. I. 




































































































PI. //. 
























































































/'/ /// 



'>•) •) 


+ ,;v/ ts> 




-7y^777^r~; 


In _ 


I 

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